Equilibrium index
You are encouraged to solve this task according to the task description, using any language you may know.
An equilibrium index of a sequence is an index into the sequence such that the sum of elements at lower indices is equal to the sum of elements at higher indices.
For example, in a sequence :
3 is an equilibrium index, because:
6 is also an equilibrium index, because:
(sum of zero elements is zero)
7 is not an equilibrium index, because it is not a valid index of sequence .
- Task;
Write a function that, given a sequence, returns its equilibrium indices (if any).
Assume that the sequence may be very long.
F eqindex(arr)
R (0 .< arr.len).filter(i -> sum(@arr[0.<i]) == sum(@arr[i+1..]))
print(eqindex([-7, 1, 5, 2, -4, 3, 0]))- Output:
[3, 6]
REPORT equilibrium_index.
TYPES: y_i TYPE STANDARD TABLE OF i WITH EMPTY KEY.
cl_demo_output=>display( REDUCE y_i( LET sequences = VALUE y_i( ( -7 ) ( 1 ) ( 5 ) ( 2 ) ( -4 ) ( 3 ) ( 0 ) )
total_sum = REDUCE #( INIT sum = 0
FOR sequence IN sequences
NEXT sum = sum + ( sequence ) ) IN
INIT x = VALUE y_i( )
y = 0
FOR i = 1 UNTIL i > lines( sequences )
LET z = sequences[ i ] IN
NEXT x = COND #( WHEN y = ( total_sum - y - z ) THEN VALUE y_i( BASE x ( i - 1 ) ) ELSE x )
y = y + z ) ).
PROC PrintArray(INT ARRAY a INT size)
INT i
Put('[)
FOR i=0 TO size-1
DO
IF i>0 THEN Put(' ) FI
PrintI(a(i))
OD
Put(']) PutE()
RETURN
INT FUNC SumRange(INT ARRAY a INT first,last)
INT sum
INT i
sum=0
FOR i=first TO last
DO
sum==+a(i)
OD
RETURN(sum)
PROC EquilibriumIndices(INT ARRAY a INT size
INT ARRAY indices INT POINTER indSize)
INT i,left,right
indSize^=0
FOR i=0 TO size-1
DO
left=SumRange(a,0,i-1)
right=SumRange(a,i+1,size-1)
IF left=right THEN
indices(indSize^)=i
indSize^==+1
FI
OD
RETURN
PROC Test(INT ARRAY a INT size)
INT ARRAY indices(100)
INT indSize
EquilibriumIndices(a,size,indices,@indSize)
Print("Array=") PrintArray(a,size)
Print("Equilibrium indices=") PrintArray(indices,indSize)
PutE()
RETURN
PROC Main()
INT ARRAY a=[65529 1 5 2 65532 3 0]
INT ARRAY b=[65535 1 65535 1 65535 1 65535]
INT ARRAY c=[1 2 3 4 5 6 7 8 9]
INT ARRAY d=[0]
Test(a,7)
Test(b,7)
Test(c,9)
Test(d,1)
RETURN- Output:
Screenshot from Atari 8-bit computer
Array=[-7 1 5 2 -4 3 0] Equilibrium indices=[3 6] Array=[-1 1 -1 1 -1 1 -1] Equilibrium indices=[0 1 2 3 4 5 6] Array=[1 2 3 4 5 6 7 8 9] Equilibrium indices=[] Array=[0] Equilibrium indices=[0]
Generic solution that returns a Vector of Indices.
equilibrium.ads:
with Ada.Containers.Vectors;
generic
type Index_Type is range <>;
type Element_Type is private;
Zero : Element_Type;
with function "+" (Left, Right : Element_Type) return Element_Type is <>;
with function "-" (Left, Right : Element_Type) return Element_Type is <>;
with function "=" (Left, Right : Element_Type) return Boolean is <>;
type Array_Type is private;
with function Element (From : Array_Type; Key : Index_Type) return Element_Type is <>;
package Equilibrium is
package Index_Vectors is new Ada.Containers.Vectors
(Index_Type => Positive, Element_Type => Index_Type);
function Get_Indices (From : Array_Type) return Index_Vectors.Vector;
end Equilibrium;
equilibrium.adb:
package body Equilibrium is
function Get_Indices (From : Array_Type) return Index_Vectors.Vector is
Result : Index_Vectors.Vector;
Right_Sum, Left_Sum : Element_Type := Zero;
begin
for Index in Index_Type'Range loop
Right_Sum := Right_Sum + Element (From, Index);
end loop;
for Index in Index_Type'Range loop
Right_Sum := Right_Sum - Element (From, Index);
if Left_Sum = Right_Sum then
Index_Vectors.Append (Result, Index);
end if;
Left_Sum := Left_Sum + Element (From, Index);
end loop;
return Result;
end Get_Indices;
end Equilibrium;
Test program using two different versions, one with vectors and one with arrays:
with Ada.Text_IO;
with Equilibrium;
with Ada.Containers.Vectors;
procedure Main is
subtype Index_Type is Positive range 1 .. 7;
package Vectors is new Ada.Containers.Vectors
(Element_Type => Integer, Index_Type => Index_Type);
type Plain_Array is array (Index_Type) of Integer;
function Element (From : Plain_Array; Key : Index_Type) return Integer is
begin
return From (Key);
end Element;
package Vector_Equilibrium is new Equilibrium
(Index_Type => Index_Type,
Element_Type => Integer,
Zero => 0,
Array_Type => Vectors.Vector,
Element => Vectors.Element);
package Array_Equilibrium is new Equilibrium
(Index_Type => Index_Type,
Element_Type => Integer,
Zero => 0,
Array_Type => Plain_Array);
My_Vector : Vectors.Vector;
My_Array : Plain_Array := (-7, 1, 5, 2, -4, 3, 0);
Vector_Result : Vector_Equilibrium.Index_Vectors.Vector;
Array_Result : Array_Equilibrium.Index_Vectors.Vector :=
Array_Equilibrium.Get_Indices (My_Array);
begin
Vectors.Append (My_Vector, -7);
Vectors.Append (My_Vector, 1);
Vectors.Append (My_Vector, 5);
Vectors.Append (My_Vector, 2);
Vectors.Append (My_Vector, -4);
Vectors.Append (My_Vector, 3);
Vectors.Append (My_Vector, 0);
Vector_Result := Vector_Equilibrium.Get_Indices (My_Vector);
Ada.Text_IO.Put_Line ("Results:");
Ada.Text_IO.Put ("Array: ");
for I in Array_Result.First_Index .. Array_Result.Last_Index loop
Ada.Text_IO.Put (Integer'Image (Array_Equilibrium.Index_Vectors.Element (Array_Result, I)));
end loop;
Ada.Text_IO.New_Line;
Ada.Text_IO.Put ("Vector: ");
for I in Vector_Result.First_Index .. Vector_Result.Last_Index loop
Ada.Text_IO.Put (Integer'Image (Vector_Equilibrium.Index_Vectors.Element (Vector_Result, I)));
end loop;
Ada.Text_IO.New_Line;
end Main;
- Output:
(Index_Type is based on 1):
Results: Array: 4 7 Vector: 4 7
Version that works with Ada 95, too:
equilibrium.adb:
with Ada.Text_IO;
procedure Equilibrium is
type Integer_Sequence is array (Positive range <>) of Integer;
function Seq_Img (From : Integer_Sequence) return String is
begin
if From'First /= From'Last then
return " " & Integer'Image (From (From'First)) &
Seq_Img (From (From'First + 1 .. From'Last));
else
return " " & Integer'Image (From (From'First));
end if;
end Seq_Img;
type Boolean_Sequence is array (Positive range <>) of Boolean;
function Seq_Img (From : Boolean_Sequence) return String is
begin
if From'First > From'Last then
return "";
end if;
if From (From'First) then
return Integer'Image (From'First) &
Seq_Img (From (From'First + 1 .. From'Last));
else
return Seq_Img (From (From'First + 1 .. From'Last));
end if;
end Seq_Img;
function Get_Indices (From : Integer_Sequence) return Boolean_Sequence is
Result : Boolean_Sequence (From'Range) := (others => False);
Left_Sum, Right_Sum : Integer := 0;
begin
for Index in From'Range loop
Right_Sum := Right_Sum + From (Index);
end loop;
for Index in From'Range loop
Right_Sum := Right_Sum - From (Index);
Result (Index) := Left_Sum = Right_Sum;
Left_Sum := Left_Sum + From (Index);
end loop;
return Result;
end Get_Indices;
X1 : Integer_Sequence := (-7, 1, 5, 2, -4, 3, 0);
X1_Result : Boolean_Sequence := Get_Indices (X1);
X2 : Integer_Sequence := ( 2, 4, 6);
X2_Result : Boolean_Sequence := Get_Indices (X2);
X3 : Integer_Sequence := ( 2, 9, 2);
X3_Result : Boolean_Sequence := Get_Indices (X3);
X4 : Integer_Sequence := ( 1, -1, 1, -1, 1 ,-1, 1);
X4_Result : Boolean_Sequence := Get_Indices (X4);
begin
Ada.Text_IO.Put_Line ("Results:");
Ada.Text_IO.New_Line;
Ada.Text_IO.Put_Line ("X1:" & Seq_Img (X1));
Ada.Text_IO.Put_Line ("Eqs:" & Seq_Img (X1_Result));
Ada.Text_IO.New_Line;
Ada.Text_IO.Put_Line ("X2:" & Seq_Img (X2));
Ada.Text_IO.Put_Line ("Eqs:" & Seq_Img (X2_Result));
Ada.Text_IO.New_Line;
Ada.Text_IO.Put_Line ("X3:" & Seq_Img (X3));
Ada.Text_IO.Put_Line ("Eqs:" & Seq_Img (X3_Result));
Ada.Text_IO.New_Line;
Ada.Text_IO.Put_Line ("X4:" & Seq_Img (X4));
Ada.Text_IO.Put_Line ("Eqs:" & Seq_Img (X4_Result));
end Equilibrium;
- Output:
Results: X1: -7 1 5 2 -4 3 0 Eqs: 4 7 X2: 2 4 6 Eqs: X3: 2 9 2 Eqs: 2 X4: 1 -1 1 -1 1 -1 1 Eqs: 1 2 3 4 5 6 7
list
eqindex(list l)
{
integer e, i, s, sum;
list x;
s = sum = 0;
l.ucall(add_i, 1, sum);
for (i, e in l) {
if (s * 2 + e == sum) {
x.append(i);
}
s += e;
}
x;
}
integer
main(void)
{
list(-7, 1, 5, 2, -4, 3, 0).eqindex.ucall(o_, 0, "\n");
0;
}COMMENT Equilibrium index;
BEGIN
PROCEDURE writearray(low, up, nums, nl);
VALUE low, up, nl;
INTEGER low, up;
INTEGER ARRAY nums;
BOOLEAN nl;
BEGIN
INTEGER i;
FOR i := low STEP 1 UNTIL up DO
write(1, nums[i]);
IF nl THEN
skip(1)
END;
PROCEDURE writeindicesoftrue(low, up, bools, nl);
VALUE low, up, nl;
INTEGER low, up;
BOOLEAN ARRAY bools;
BOOLEAN nl;
BEGIN
INTEGER i;
FOR i := low STEP 1 UNTIL up DO
IF bools[i] THEN
write(1, i);
IF nl THEN
skip(1)
END;
PROCEDURE getindices(low, up, nums, result);
VALUE low, up; INTEGER low, up;
INTEGER ARRAY nums;
BOOLEAN ARRAY result;
BEGIN
INTEGER leftsum, rightsum;
INTEGER i;
leftsum := 0;
rightsum := 0;
FOR i := low STEP 1 UNTIL up DO
rightsum := rightsum + nums[i];
FOR i := low STEP 1 UNTIL up DO
BEGIN
rightsum := rightsum - nums[i];
result[i] := (leftsum = rightsum);
leftsum := leftsum + nums[i]
END;
END;
INTEGER ARRAY x1[0 : 6];
BOOLEAN ARRAY x1result[0 : 6];
INTEGER ARRAY x2[0 : 2];
BOOLEAN ARRAY x2result[0 : 2];
INTEGER ARRAY x3[0 : 2];
BOOLEAN ARRAY x3result[0 : 2];
INTEGER ARRAY x4[0 : 6];
BOOLEAN ARRAY x4result[0 : 6];
x1[0] := -7; x1[1] := 1; x1[2] := 5; x1[3] := 2;
x1[4] := -4; x1[5] := 3; x1[6] := 0;
getindices(0, 6, x1, x1result);
x2[0] := 2; x2[1] := 4; x2[2] := 6;
getindices(0, 2, x2, x2result);
x3[0] := 2; x3[1] := 9; x3[2] := 2;
getindices(0, 2, x3, x3result);
x4[0] := 1; x4[1] := -1; x4[2] := 1; x4[3] := -1;
x4[4] := 1; x4[5] := -1; x4[6] := 1;
getindices(0, 6, x4, x4result);
text(1, "Results:");
skip(1);
skip(1);
text(1, "X1: ");
writearray(0, 6, x1, TRUE);
text(1, "Eqs: ");
writeindicesoftrue(0, 6, x1result, TRUE);
skip(1);
text(1, "X2: ");
writearray(0, 2, x2, TRUE);
text(1, "Eqs: ");
writeindicesoftrue(0, 2, x2result, TRUE);
skip(1);
text(1, "X3: ");
writearray(0, 2, x3, TRUE);
text(1, "Eqs: ");
writeindicesoftrue(0, 2, x3result, TRUE);
skip(1);
text(1, "X4: ");
writearray(0, 6, x4, TRUE);
text(1, "Eqs: ");
writeindicesoftrue(0, 6, x4result, TRUE);
END FINISH- Output:
Results: X1: -7 1 5 2 -4 3 0 Eqs: 3 6 X2: 2 4 6 Eqs: X3: 2 9 2 Eqs: 1 X4: 1 -1 1 -1 1 -1 1 Eqs: 0 1 2 3 4 5 6
MODE YIELDINT = PROC(INT)VOID;
PROC gen equilibrium index = ([]INT arr, YIELDINT yield)VOID:
(
INT sum := 0;
FOR i FROM LWB arr TO UPB arr DO
sum +:= arr[i]
OD;
INT left:=0, right:=sum;
FOR i FROM LWB arr TO UPB arr DO
right -:= arr[i];
IF left = right THEN yield(i) FI;
left +:= arr[i]
OD
);
test:(
[]INT arr = []INT(-7, 1, 5, 2, -4, 3, 0)[@0];
# FOR INT index IN # gen equilibrium index(arr, # ) DO ( #
## (INT index)VOID:
print(index)
# OD # );
print(new line)
)- Output:
+3 +6
Functional
(ES6 version)
-- equilibriumIndices :: [Int] -> [Int]
on equilibriumIndices(xs)
script balancedPair
on |λ|(a, pair, i)
set {x, y} to pair
if x = y then
{i - 1} & a
else
a
end if
end |λ|
end script
script plus
on |λ|(a, b)
a + b
end |λ|
end script
-- Fold over zipped pairs of sums from left
-- and sums from right
foldr(balancedPair, {}, ¬
zip(scanl1(plus, xs), scanr1(plus, xs)))
end equilibriumIndices
-- TEST -----------------------------------------------------------------------
on run
map(equilibriumIndices, {¬
{-7, 1, 5, 2, -4, 3, 0}, ¬
{2, 4, 6}, ¬
{2, 9, 2}, ¬
{1, -1, 1, -1, 1, -1, 1}, ¬
{1}, ¬
{}})
--> {{3, 6}, {}, {1}, {0, 1, 2, 3, 4, 5, 6}, {0}, {}}
end run
-- GENERIC FUNCTIONS ----------------------------------------------------------
-- foldl :: (a -> b -> a) -> a -> [b] -> a
on foldl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldl
-- foldr :: (a -> b -> a) -> a -> [b] -> a
on foldr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
repeat with i from lng to 1 by -1
set v to |λ|(v, item i of xs, i, xs)
end repeat
return v
end tell
end foldr
-- init :: [a] -> [a]
on init(xs)
set lng to length of xs
if lng > 1 then
items 1 thru -2 of xs
else if lng > 0 then
{}
else
missing value
end if
end init
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- min :: Ord a => a -> a -> a
on min(x, y)
if y < x then
y
else
x
end if
end min
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- scanl :: (b -> a -> b) -> b -> [a] -> [b]
on scanl(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
set lst to {startValue}
repeat with i from 1 to lng
set v to |λ|(v, item i of xs, i, xs)
set end of lst to v
end repeat
return lst
end tell
end scanl
-- scanl1 :: (a -> a -> a) -> [a] -> [a]
on scanl1(f, xs)
if length of xs > 0 then
scanl(f, item 1 of xs, tail(xs))
else
{}
end if
end scanl1
-- scanr :: (b -> a -> b) -> b -> [a] -> [b]
on scanr(f, startValue, xs)
tell mReturn(f)
set v to startValue
set lng to length of xs
set lst to {startValue}
repeat with i from lng to 1 by -1
set v to |λ|(v, item i of xs, i, xs)
set end of lst to v
end repeat
return reverse of lst
end tell
end scanr
-- scanr1 :: (a -> a -> a) -> [a] -> [a]
on scanr1(f, xs)
if length of xs > 0 then
scanr(f, item -1 of xs, init(xs))
else
{}
end if
end scanr1
-- tail :: [a] -> [a]
on tail(xs)
if length of xs > 1 then
items 2 thru -1 of xs
else
{}
end if
end tail
-- zip :: [a] -> [b] -> [(a, b)]
on zip(xs, ys)
set lng to min(length of xs, length of ys)
set lst to {}
repeat with i from 1 to lng
set end of lst to {item i of xs, item i of ys}
end repeat
return lst
end zip
- Output:
{{3, 6}, {}, {1}, {0, 1, 2, 3, 4, 5, 6}, {0}, {}}
Straightforward
on equilibriumIndices(sequence)
script o
property seq : sequence
property output : {}
end script
set loSum to 0
set hiSum to 0
repeat with value in o's seq
set hiSum to hiSum + value
end repeat
repeat with i from 1 to (count o's seq)
set value to o's seq's item i
set hiSum to hiSum - value
if (hiSum = loSum) then set o's output's end to i
set loSum to loSum + value
end repeat
return o's output
end equilibriumIndices
equilibriumIndices({-7, 1, 5, 2, -4, 3, 0})
- Output:
AppleScript uses 1-based indices.
{4, 7}
eqIndex: function [row][
suml: 0
delayed: 0
sumr: sum row
result: []
loop.with:'i row 'r [
suml: suml + delayed
sumr: sumr - r
delayed: r
if suml = sumr -> 'result ++ i
]
return result
]
data: @[
@[neg 7, 1, 5, 2, neg 4, 3, 0]
@[2 4 6]
@[2 9 2]
@[1 neg 1 1 neg 1 1 neg 1 1]
]
loop data 'd ->
print [pad.right join.with:", " to [:string] d 25 "=> equilibrium index:" eqIndex d]
- Output:
-7, 1, 5, 2, -4, 3, 0 => equilibrium index: [3 6] 2, 4, 6 => equilibrium index: [] 2, 9, 2 => equilibrium index: [1] 1, -1, 1, -1, 1, -1, 1 => equilibrium index: [0 1 2 3 4 5 6]
Equilibrium_index(list, BaseIndex=0){
StringSplit, A, list, `,
Loop % A0 {
i := A_Index , Pre := Post := 0
loop, % A0
if (A_Index < i)
Pre += A%A_Index%
else if (A_Index > i)
Post += A%A_Index%
if (Pre = Post)
Res .= (Res?", ":"") i - (BaseIndex?0:1)
}
return Res
}
Examples:
list = -7, 1, 5, 2, -4, 3, 0
MsgBox % Equilibrium_index(list)
- Output:
3, 6
# syntax: GAWK -f EQUILIBRIUM_INDEX.AWK
BEGIN {
main("-7 1 5 2 -4 3 0")
main("2 4 6")
main("2 9 2")
main("1 -1 1 -1 1 -1 1")
exit(0)
}
function main(numbers, x) {
x = equilibrium(numbers)
printf("numbers: %s\n",numbers)
printf("indices: %s\n\n",length(x)==0?"none":x)
}
function equilibrium(numbers, arr,i,leftsum,leng,str,sum) {
leng = split(numbers,arr," ")
for (i=1; i<=leng; i++) {
sum += arr[i]
}
for (i=1; i<=leng; i++) {
sum -= arr[i]
if (leftsum == sum) {
str = str i " "
}
leftsum += arr[i]
}
return(str)
}
Output:
numbers: -7 1 5 2 -4 3 0 indices: 4 7 numbers: 2 4 6 indices: none numbers: 2 9 2 indices: 2 numbers: 1 -1 1 -1 1 -1 1 indices: 1 2 3 4 5 6 7
See also Visual Basic.
Indices from 0, like in the task.
100 REM Equilibrium index
110 DECLARE EXTERNAL SUB PrintArray
120 DECLARE EXTERNAL SUB PrintIndicesOfTrue
130 DECLARE EXTERNAL SUB GetIndices
140 DIM X1(0 TO 6), X1Result(0 TO 6)
150 FOR I = 0 TO 6
160 READ X1(I)
170 NEXT I
180 DATA -7, 1, 5, 2, -4, 3, 0
190 CALL GetIndices(X1, X1Result)
200 DIM X2(0 TO 2), X2Result(0 TO 2)
210 FOR I = 0 TO 2
220 READ X2(I)
230 NEXT I
240 DATA 2, 4, 6
250 CALL GetIndices(X2, X2Result)
260 DIM X3(0 TO 2), X3Result(0 TO 2)
270 FOR I = 0 TO 2
280 READ X3(I)
290 NEXT I
300 DATA 2, 9, 2
310 CALL GetIndices(X3, X3Result)
320 DIM X4(0 TO 6), X4Result(0 TO 6)
330 FOR I = 0 TO 6
340 READ X4(I)
350 NEXT I
360 DATA 1, -1, 1, -1, 1 ,-1, 1
370 CALL GetIndices(X4, X4Result)
380 PRINT "Results:"
390 PRINT
400 PRINT "X1:";
410 CALL PrintArray(X1, 1)
420 PRINT "Eqs:";
430 CALL PrintIndicesOfTrue(X1Result, 1)
440 PRINT
450 PRINT "X2:";
460 CALL PrintArray(X2, 1)
470 PRINT "Eqs:";
480 CALL PrintIndicesOfTrue(X2Result, 1)
490 PRINT
500 PRINT "X3:";
510 CALL PrintArray(X3, 1)
520 PRINT "Eqs:";
530 CALL PrintIndicesOfTrue(X3Result, 1)
540 PRINT
550 PRINT "X4:";
560 CALL PrintArray(X4, 1)
570 PRINT "Eqs:";
580 CALL PrintIndicesOfTrue(X4Result, 1)
590 END
600 !
610 EXTERNAL SUB PrintArray(Nums(), NL)
620 FOR I = LBOUND(Nums) TO UBOUND(Nums)
630 PRINT Nums(I);
640 NEXT I
650 IF NL <> 0 THEN PRINT
660 END SUB
670 !
680 EXTERNAL SUB PrintIndicesOfTrue(Bools(), NL)
690 FOR I = LBOUND(Bools) TO UBOUND(Bools)
700 IF Bools(I) <> 0 THEN PRINT I;
710 NEXT I
720 IF NL <> 0 THEN PRINT
730 END SUB
740 !
750 EXTERNAL SUB GetIndices(Nums(), Result())
760 IF (LBOUND(Nums) <> LBOUND(Result)) OR (UBOUND(Nums) <> UBOUND(Result)) THEN
770 PRINT "Range of indices in both arrays should be the same"
780 STOP
790 END IF
800 LET LeftSum = 0
810 LET RightSum = 0
820 FOR I = LBOUND(Nums) TO UBOUND(Nums)
830 LET RightSum = RightSum + Nums (I)
840 NEXT I
850 FOR I = LBOUND(Nums) TO UBOUND(Nums)
860 LET RightSum = RightSum - Nums (I)
870 IF LeftSum = RightSum THEN
880 LET Result(I) = 1
890 ELSE
900 LET Result(I) = 0
910 END IF
920 LET LeftSum = LeftSum + Nums(I)
930 NEXT I
940 END SUB
- Output:
Results: X1:-7 1 5 2 -4 3 0 Eqs: 3 6 X2: 2 4 6 Eqs: X3: 2 9 2 Eqs: 1 X4: 1 -1 1 -1 1 -1 1 Eqs: 0 1 2 3 4 5 6
arraybase 1
dim list = {-7, 1, 5, 2, -4, 3, 0}
print "equilibrium indices are : "; equilibrium(list)
end
function equilibrium (l)
r = 0: s = 0
e$ = ""
for n = 1 to l[?]
s += l[n]
next
for i = 1 to l[?]
if r = s - r - l[i] then e$ += string(i-1) + " "
r += l[i]
next
e$ = left(e$, length(e$)-1)
return e$
end function
- Output:
The equilibrium indices are : 3 6
BBC BASIC's SUM function is useful for this task.
DIM list(6)
list() = -7, 1, 5, 2, -4, 3, 0
PRINT "Equilibrium indices are " FNequilibrium(list())
END
DEF FNequilibrium(l())
LOCAL i%, r, s, e$
s = SUM(l())
FOR i% = 0 TO DIM(l(),1)
IF r = s - r - l(i%) THEN e$ += STR$(i%) + ","
r += l(i%)
NEXT
= LEFT$(e$)
Output:
Equilibrium indices are 3,6
' FB 1.05.0 Win64
Sub equilibriumIndices (a() As Integer, b() As Integer)
If UBound(a) = -1 Then Return '' empty array
Dim sum As Integer = 0
Dim count As Integer = 0
For i As Integer = LBound(a) To UBound(a) : sum += a(i) : Next
Dim sumLeft As Integer = 0, sumRight As Integer = 0
For i As Integer = LBound(a) To UBound(a)
sumRight = sum - sumLeft - a(i)
If sumLeft = sumRight Then
Redim Preserve b(0 To Count)
b(count) = i
count += 1
End If
sumLeft += a(i)
Next
End Sub
Dim a(0 To 6) As Integer = { -7, 1, 5, 2, -4, 3, 0 }
Dim b() As Integer
equilibriumIndices a(), b()
If UBound(b) = -1 Then
Print "There are no equilibrium indices"
ElseIf UBound(b) = LBound(b) Then
Print "The only equilibrium index is : "; b(LBound(b))
Else
Print "The equilibrium indices are : "
For i As Integer = LBound(b) To UBound(b) : Print b(i); " "; : Next
End If
Print
Print "Press any key to quit"
Sleep
- Output:
The equilibrium indices are : 3 6
a(0)=-7
a(1)=1
a(2)=5
a(3)=2
a(4)=-4
a(5)=3
a(6)=0
print "EQ Indices are ";EQindex$("a",0,6)
wait
function EQindex$(b$,mini,maxi)
if mini>=maxi then exit function
sum=0
for i = mini to maxi
sum=sum+eval(b$;"(";i;")")
next
sumA=0:sumB=sum
for i = mini to maxi
sumB = sumB - eval(b$;"(";i;")")
if sumA=sumB then EQindex$=EQindex$+str$(i)+", "
sumA = sumA + eval(b$;"(";i;")")
next
if len(EQindex$)>0 then EQindex$=mid$(EQindex$, 1, len(EQindex$)-2) 'remove last ", "
end function- Output:
EQ Indices are 3, 6
If OpenConsole()
Define i, c=CountProgramParameters()-1
For i=0 To c
Define j, LSum=0, RSum=0
For j=0 To c
If j<i
LSum+Val(ProgramParameter(j))
ElseIf j>i
RSum+Val(ProgramParameter(j))
EndIf
Next j
If LSum=RSum: PrintN(Str(i)): EndIf
Next i
EndIf
- Output:
> Equilibrium.exe -7 1 5 2 -4 3 0 3 6
The QuickBASIC solution works without any changes.
The QuickBASIC solution works without any changes.
' Equilibrium index
DECLARE SUB PrintArray (Nums() AS INTEGER, NL AS INTEGER)
DECLARE SUB PrintIndicesOfTrue (Bools() AS INTEGER, NL AS INTEGER)
DECLARE SUB GetIndices (Nums() AS INTEGER, Result() AS INTEGER)
DIM I AS INTEGER
DIM X1(0 TO 6) AS INTEGER, X1Result(0 TO 6) AS INTEGER
FOR I = 0 TO 6
READ X1(I)
NEXT I
DATA -7, 1, 5, 2, -4, 3, 0
CALL GetIndices(X1(), X1Result())
DIM X2(0 TO 2) AS INTEGER, X2Result(0 TO 2) AS INTEGER
FOR I = 0 TO 2
READ X2(I)
NEXT I
DATA 2, 4, 6
CALL GetIndices(X2(), X2Result())
DIM X3(0 TO 2) AS INTEGER, X3Result(0 TO 2) AS INTEGER
FOR I = 0 TO 2
READ X3(I)
NEXT I
DATA 2, 9, 2
CALL GetIndices(X3(), X3Result())
DIM X4(0 TO 6) AS INTEGER, X4Result(0 TO 6) AS INTEGER
FOR I = 0 TO 6
READ X4(I)
NEXT I
DATA 1, -1, 1, -1, 1 ,-1, 1
CALL GetIndices(X4(), X4Result())
PRINT "Results:"
PRINT
PRINT "X1:";
CALL PrintArray(X1(), 1)
PRINT "Eqs:";
CALL PrintIndicesOfTrue(X1Result(), 1)
PRINT
PRINT "X2:";
CALL PrintArray(X2(), 1)
PRINT "Eqs:";
CALL PrintIndicesOfTrue(X2Result(), 1)
PRINT
PRINT "X3:";
CALL PrintArray(X3(), 1)
PRINT "Eqs:";
CALL PrintIndicesOfTrue(X3Result(), 1)
PRINT
PRINT "X4:";
CALL PrintArray(X4(), 1)
PRINT "Eqs:";
CALL PrintIndicesOfTrue(X4Result(), 1)
END
SUB GetIndices (Nums() AS INTEGER, Result() AS INTEGER)
IF (LBOUND(Nums) <> LBOUND(Result)) OR (UBOUND(Nums) <> UBOUND(Result)) THEN
PRINT "Range of indices in both arrays should be the same"
STOP
END IF
DIM LeftSum AS INTEGER, RightSum AS INTEGER
DIM I AS INTEGER
LeftSum = 0
RightSum = 0
FOR I = LBOUND(Nums) TO UBOUND(Nums)
RightSum = RightSum + Nums(I)
NEXT I
FOR I = LBOUND(Nums) TO UBOUND(Nums)
RightSum = RightSum - Nums(I)
Result(I) = (LeftSum = RightSum)
LeftSum = LeftSum + Nums(I)
NEXT I
END SUB
SUB PrintArray (Nums() AS INTEGER, NL AS INTEGER)
FOR I = LBOUND(Nums) TO UBOUND(Nums)
PRINT Nums(I);
NEXT I
IF NL THEN PRINT
END SUB
SUB PrintIndicesOfTrue (Bools() AS INTEGER, NL AS INTEGER)
FOR I = LBOUND(Bools) TO UBOUND(Bools)
IF Bools(I) THEN PRINT I;
NEXT I
IF NL THEN PRINT
END SUB
- Output:
X1:-7 1 5 2 -4 3 0 Eqs: 3 6 X2: 2 4 6 Eqs: X3: 2 9 2 Eqs: 1 X4: 1 -1 1 -1 1 -1 1 Eqs: 0 1 2 3 4 5 6
10 DATA 7,-7,1,5,2,-4,3,0
20 READ n
30 DIM a(n): LET sum=0: LET leftsum=0: LET s$=""
40 FOR i=1 TO n: READ a(i): LET sum=sum+a(i): NEXT i
50 FOR i=1 TO n
60 LET sum=sum-a(i)
70 IF leftsum=sum THEN LET s$=s$+STR$ i+" "
80 LET leftsum=leftsum+a(i)
90 NEXT i
100 PRINT "Numbers: ";
110 FOR i=1 TO n: PRINT a(i);" ";: NEXT i
120 PRINT '"Indices: ";s$
@echo off
setlocal enabledelayedexpansion
call :equilibrium-index "-7 1 5 2 -4 3 0"
call :equilibrium-index "2 4 6"
call :equilibrium-index "2 9 2"
call :equilibrium-index "1 -1 1 -1 1 -1 1"
pause>nul
exit /b
%== The Function ==%
:equilibrium-index <sequence with quotes>
::Set the pseudo-array sequence...
set "seq=%~1"
set seq.length=0
for %%S in (!seq!) do (
set seq[!seq.length!]=%%S
set /a seq.length+=1
)
::Initialization of other variables...
set "equilms="
set /a last=seq.length - 1
::The main checking...
for /l %%e in (0,1,!last!) do (
set left=0
set right=0
for /l %%i in (0,1,!last!) do (
if %%i lss %%e (set /a left+=!seq[%%i]!)
if %%i gtr %%e (set /a right+=!seq[%%i]!)
)
if !left!==!right! (
if defined equilms (
set "equilms=!equilms! %%e"
) else (
set "equilms=%%e"
)
)
)
echo [!equilms!]
goto :EOF
%==/The Function ==%
- Output:
[3 6] [] [1] [0 1 2 3 4 5 6]
#include <stdio.h>
#include <stdlib.h>
int list[] = {-7, 1, 5, 2, -4, 3, 0};
int eq_idx(int *a, int len, int **ret)
{
int i, sum, s, cnt;
/* alloc long enough: if we can afford the original list,
* we should be able to afford to this. Beats a potential
* million realloc() calls. Even if memory is a real concern,
* there's no garantee the result is shorter than the input anyway */
cnt = s = sum = 0;
*ret = malloc(sizeof(int) * len);
for (i = 0; i < len; i++)
sum += a[i];
for (i = 0; i < len; i++) {
if (s * 2 + a[i] == sum) {
(*ret)[cnt] = i;
cnt++;
}
s += a[i];
}
/* uncouraged way to use realloc since it can leak memory, for example */
*ret = realloc(*ret, cnt * sizeof(int));
return cnt;
}
int main()
{
int i, cnt, *idx;
cnt = eq_idx(list, sizeof(list) / sizeof(int), &idx);
printf("Found:");
for (i = 0; i < cnt; i++)
printf(" %d", idx[i]);
printf("\n");
return 0;
}
using System;
using System.Collections.Generic;
using System.Linq;
class Program
{
static IEnumerable<int> EquilibriumIndices(IEnumerable<int> sequence)
{
var left = 0;
var right = sequence.Sum();
var index = 0;
foreach (var element in sequence)
{
right -= element;
if (left == right)
{
yield return index;
}
left += element;
index++;
}
}
static void Main()
{
foreach (var index in EquilibriumIndices(new[] { -7, 1, 5, 2, -4, 3, 0 }))
{
Console.WriteLine(index);
}
}
}
- Output:
3
6
#include <algorithm>
#include <iostream>
#include <numeric>
#include <vector>
template <typename T>
std::vector<size_t> equilibrium(T first, T last)
{
typedef typename std::iterator_traits<T>::value_type value_t;
value_t left = 0;
value_t right = std::accumulate(first, last, value_t(0));
std::vector<size_t> result;
for (size_t index = 0; first != last; ++first, ++index)
{
right -= *first;
if (left == right)
{
result.push_back(index);
}
left += *first;
}
return result;
}
template <typename T>
void print(const T& value)
{
std::cout << value << "\n";
}
int main()
{
const int data[] = { -7, 1, 5, 2, -4, 3, 0 };
std::vector<size_t> indices(equilibrium(data, data + 7));
std::for_each(indices.begin(), indices.end(), print<size_t>);
}
- Output:
3 6
(defn equilibrium [lst]
(loop [acc '(), i 0, left 0, right (apply + lst), lst lst]
(if (empty? lst)
(reverse acc)
(let [[x & xs] lst
right (- right x)
acc (if (= left right) (cons i acc) acc)]
(recur acc (inc i) (+ left x) right xs)))))
- Output:
> (equilibrium [-7, 1, 5, 2, -4, 3, 0]) (3 6)
(defun dflt-on-nil (v dflt)
(if v v dflt))
(defun eq-index (v)
(do*
((stack nil)
(i 0 (+ 1 i))
(rest v (cdr rest))
(lsum 0)
(rsum (apply #'+ (cdr v))))
;; Reverse here is not strictly necessary
((null rest) (reverse stack))
(if (eql lsum rsum) (push i stack))
(setf lsum (+ lsum (car rest)))
(setf rsum (- rsum (dflt-on-nil (cadr rest) 0)))))
- Output:
(eq-index '(-7 1 5 2 -4 3 0))
(3 6)
def equilibrium_indices (arr)
left = 0
right = arr.sum
indices = [] of Int32
(0...arr.size).each do |i|
right -= arr[i]
indices << i if left == right
left += arr[i]
end
indices
end
[[-7, 1, 5, 2, -4, 3, 0], [2, 4, 6], [2, 9, 2], [1,-1, 1,-1, 1,-1, 1]].each do |arr|
print arr, ": ", equilibrium_indices(arr), "\n"
end
- Output:
[-7, 1, 5, 2, -4, 3, 0]: [3, 6] [2, 4, 6]: [] [2, 9, 2]: [1] [1, -1, 1, -1, 1, -1, 1]: [0, 1, 2, 3, 4, 5, 6]
More Functional Style
import std.stdio, std.algorithm, std.range, std.functional;
auto equilibrium(Range)(Range r) pure nothrow @safe /*@nogc*/ {
return r.length.iota.filter!(i => r[0 .. i].sum == r[i + 1 .. $].sum);
}
void main() {
[-7, 1, 5, 2, -4, 3, 0].equilibrium.writeln;
}
- Output:
[3, 6]
Less Functional Style
Same output.
import std.stdio, std.algorithm;
size_t[] equilibrium(T)(in T[] items) @safe pure nothrow {
size_t[] result;
T left = 0, right = items.sum;
foreach (immutable i, e; items) {
right -= e;
if (right == left)
result ~= i;
left += e;
}
return result;
}
void main() {
[-7, 1, 5, 2, -4, 3, 0].equilibrium.writeln;
}
See Pascal.
DuckDB lists and arrays have an index origin of 1 and the solution presented below conforms with that convention. This solution is not memory-efficient as it uses a recursive Common Table Expression (CTE).
create or replace function equilibrium_indices(lst) as table (
with recursive cte as (
select 0 as ix, 0 as before, 0 as pivotor, list_sum(lst) as after
union all
select ix+1,
before+pivotor,
lst[ix+1],
after - lst[ix+1]
from cte
where ix<length(lst)
)
select ix
from cte
where before = after and ix>0
);
A typescript of examples follows, where "D " signifies the DuckDB prompt.
- Output:
D from equilibrium_indices( [-7, 1, 5, 2, -4, 3, 0] ); ┌───────┐ │ ix │ │ int32 │ ├───────┤ │ 4 │ │ 7 │ └───────┘ D select count(*) from equilibrium_indices( list_transform(range(0,10000), x -> 0) ); ┌──────────────┐ │ count_star() │ │ int64 │ ├──────────────┤ │ 10000 │ └──────────────┘ D from equilibrium_indices( list_reverse(range(0,500)) || range(0,500)) ; ┌───────┐ │ ix │ │ int32 │ ├───────┤ │ 500 │ │ 501 │ └───────┘
func[] equind a[] .
for v in a[] : sumr += v
for i to len a[]
sumr -= a[i]
if suml = sumr : r[] &= i
suml += a[i]
.
return r[]
.
print equind [ -7 1 5 2 -4 3 0 ]- Output:
[ 4 7 ]
ELENA 6.x :
import extensions;
import system'routines;
import system'collections;
import extensions'routines;
class EquilibriumEnumerator : Enumerator
{
int left;
int right;
int index;
Enumerator en;
constructor new(Enumerator en)
{
this en := en;
self.reset()
}
constructor new(Enumerable list)
<= new(list.enumerator());
constructor new(o)
<= new(cast Enumerable(o));
bool next()
{
index += 1;
while(en.next())
{
var element := *en;
right -= element;
bool found := (left == right);
left += element;
if (found)
{
^ true
};
index += 1
};
^ false
}
reset()
{
en.reset();
left := 0;
right := en.summarize();
index := -1;
en.reset();
}
get Value() = index;
enumerable() => en;
}
public Program()
{
EquilibriumEnumerator.new(new int[]{ -7, 1, 5, 2, -4, 3, 0 })
.forEach(PrintingLn)
}3 6
computes either side each time.
defmodule Equilibrium do
def index(list) do
last = length(list)
Enum.filter(0..last-1, fn i ->
Enum.sum(Enum.slice(list, 0, i)) == Enum.sum(Enum.slice(list, i+1..last))
end)
end
end
faster version:
defmodule Equilibrium do
def index(list), do: index(list,0,0,Enum.sum(list),[])
defp index([],_,_,_,acc), do: Enum.reverse(acc)
defp index([h|t],i,left,right,acc) when left==right-h, do: index(t,i+1,left+h,right-h,[i|acc])
defp index([h|t],i,left,right,acc) , do: index(t,i+1,left+h,right-h,acc)
end
Test:
indices = [
[-7, 1, 5, 2,-4, 3, 0],
[2, 4, 6],
[2, 9, 2],
[1,-1, 1,-1, 1,-1, 1]
]
Enum.each(indices, fn list ->
IO.puts "#{inspect list} => #{inspect Equilibrium.index(list)}"
end)
- Output:
[-7, 1, 5, 2, -4, 3, 0] => [3, 6] [2, 4, 6] => [] [2, 9, 2] => [1] [1, -1, 1, -1, 1, -1, 1] => [0, 1, 2, 3, 4, 5, 6]
PROGRAM EQUILIBRIUM
DIM LISTA[6]
PROCEDURE EQ(LISTA[]->RES$)
LOCAL I%,R,S,E$
FOR I%=0 TO UBOUND(LISTA,1) DO
S+=LISTA[I%]
END FOR
FOR I%=0 TO UBOUND(LISTA,1) DO
IF R=S-R-LISTA[I%] THEN E$+=STR$(I%)+"," END IF
R+=LISTA[I%]
END FOR
RES$=LEFT$(E$,LEN(E$)-1)
END PROCEDURE
BEGIN
LISTA[]=(-7,1,5,2,-4,3,0)
EQ(LISTA[]->RES$)
PRINT("Equilibrium indices are";RES$)
END PROGRAMOutput:
Equilibrium indices are 3, 6
function equilibrium(sequence s)
integer lower_sum, higher_sum
sequence indices
lower_sum = 0
higher_sum = 0
for i = 1 to length(s) do
higher_sum += s[i]
end for
indices = {}
for i = 1 to length(s) do
higher_sum -= s[i]
if lower_sum = higher_sum then
indices &= i
end if
lower_sum += s[i]
end for
return indices
end function
? equilibrium({-7,1,5,2,-4,3,0})- Output:
(Remember that indices are 1-based in Euphoria)
{4,7}
Executed in the listener. Note that accum-left and accum-right have different outputs than accumulate as they drop the final result.
USE: math.vectors
: accum-left ( seq id quot -- seq ) accumulate nip ; inline
: accum-right ( seq id quot -- seq ) [ <reversed> ] 2dip accum-left <reversed> ; inline
: equilibrium-indices ( seq -- inds )
0 [ + ] [ accum-left ] [ accum-right ] 3bi [ = ] 2map
V{ } swap dup length iota [ [ suffix ] curry [ ] if ] 2each ;
- Output:
( scratchpad ) { -7 1 5 2 -4 3 0 } equilibrium-indices .
V{ 3 6 }
Array indices are 1-based.
program Equilibrium
implicit none
integer :: array(7) = (/ -7, 1, 5, 2, -4, 3, 0 /)
call equil_index(array)
contains
subroutine equil_index(a)
integer, intent(in) :: a(:)
integer :: i
do i = 1, size(a)
if(sum(a(1:i-1)) == sum(a(i+1:size(a)))) write(*,*) i
end do
end subroutine
end program
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.
Programs in Fōrmulæ are created/edited online in its website.
In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.
Solution
In Fōrmulæ, indices are 1-based so the output of this program will be shifted up by one compared to solutions in languages with 0-based arrays.
Test cases
CFArrayRef local fn EquilibriumIndexes( arr as CFArrayRef )
NSInteger count = len(arr)
if ( count == 0 ) then return @[]
CFNumberRef sumLeft = @0, sumRight, sumAll = fn ObjectValueForKeyPath( arr, @"@sum.self" )
CFMutableArrayRef result = fn MutableArrayNew
for NSUInteger i = 0 to count - 1
CFNumberRef currVal = arr[i]
sumRight = @(dblval(sumAll) - dblval(sumLeft) - dblval(currVal))
if ( fn NumberIsEqual( sumLeft, sumRight ) )
MutableArrayAddObject( result, @(i) )
end if
sumLeft = @(dblval(sumLeft) + dblval(currVal))
next
end fn = result
void local fn DoIt
CFArrayRef arr = @[@(-7), @1, @5, @2, @(-4), @3, @0]
CFArrayRef eqIndexes = fn EquilibriumIndexes( arr )
printf @"Equilibrium indexes of [%@]: [%@]",fn ArrayComponentsJoinedByString( arr, @", " ) ,fn ArrayComponentsJoinedByString( eqIndexes, @", " )
end fn
fn DoIt
HandleEvents- Output:
Equilibrium indexes of [-7, 1, 5, 2, -4, 3, 0]: [3, 6]
package main
import (
"fmt"
"math/rand"
"time"
)
func main() {
fmt.Println(ex([]int32{-7, 1, 5, 2, -4, 3, 0}))
// sequence of 1,000,000 random numbers, with values
// chosen so that it will be likely to have a couple
// of equalibrium indexes.
rand.Seed(time.Now().UnixNano())
verylong := make([]int32, 1e6)
for i := range verylong {
verylong[i] = rand.Int31n(1001) - 500
}
fmt.Println(ex(verylong))
}
func ex(s []int32) (eq []int) {
var r, l int64
for _, el := range s {
r += int64(el)
}
for i, el := range s {
r -= int64(el)
if l == r {
eq = append(eq, i)
}
l += int64(el)
}
return
}
- Output:
[3 6] [145125 947872]
import System.Random (randomRIO)
import Data.List (findIndices, takeWhile)
import Control.Monad (replicateM)
import Control.Arrow ((&&&))
equilibr xs =
findIndices (\(a, b) -> sum a == sum b) . takeWhile (not . null . snd) $
flip ((&&&) <$> take <*> (drop . pred)) xs <$> [1 ..]
langeSliert = replicateM 2000 (randomRIO (-15, 15) :: IO Int) >>= print . equilibr
Small example
*Main> equilibr [-7, 1, 5, 2, -4, 3, 0]
[3,6]
Long random list in langeSliert (several tries for this one)
*Main> langeSliert
[231,245,259,265,382,1480,1611,1612]
Or, using default Prelude functions:
equilibriumIndices :: [Int] -> [Int]
equilibriumIndices xs =
zip3
(scanl1 (+) xs) -- Sums from the left
(scanr1 (+) xs) -- Sums from the right
[0 ..] -- Indices
>>= (\(x, y, i) -> [i | x == y])
--------------------------- TEST -------------------------
main :: IO ()
main =
mapM_
print
$ equilibriumIndices
<$> [ [-7, 1, 5, 2, -4, 3, 0],
[2, 4, 6],
[2, 9, 2],
[1, -1, 1, -1, 1, -1, 1],
[1],
[]
]
- Output:
[3,6] [] [1] [0,1,2,3,4,5,6] [0] []
- Output:
equilibrium indicies of [ -7 1 5 2 -4 3 0 ] = 4 7
equilidx=: +/\ I.@:= +/\.
- Example use:
equilidx _7 1 5 2 _4 3 0
3 6
public class Equlibrium {
public static void main(String[] args) {
int[] sequence = {-7, 1, 5, 2, -4, 3, 0};
equlibrium_indices(sequence);
}
public static void equlibrium_indices(int[] sequence){
//find total sum
int totalSum = 0;
for (int n : sequence) {
totalSum += n;
}
//compare running sum to remaining sum to find equlibrium indices
int runningSum = 0;
for (int i = 0; i < sequence.length; i++) {
int n = sequence[i];
if (totalSum - runningSum - n == runningSum) {
System.out.println(i);
}
runningSum += n;
}
}
}
- Output:
3 6
ES5
function equilibrium(a) {
var N = a.length, i, l = [], r = [], e = []
for (l[0] = a[0], r[N - 1] = a[N - 1], i = 1; i<N; i++)
l[i] = l[i - 1] + a[i], r[N - i - 1] = r[N - i] + a[N - i - 1]
for (i = 0; i < N; i++)
if (l[i] === r[i]) e.push(i)
return e
}
// test & output
[ [-7, 1, 5, 2, -4, 3, 0], // 3, 6
[2, 4, 6], // empty
[2, 9, 2], // 1
[1, -1, 1, -1, 1, -1, 1], // 0,1,2,3,4,5,6
[1], // 0
[] // empty
].forEach(function(x) {
console.log(equilibrium(x))
});
- Output:
[[3,6],[],[1],[0,1,2,3,4,5,6],[0],[]]
ES6 Procedural
Two pass O(n), returning only the first equilibrium index.
function equilibrium(arr) {
let sum = arr.reduce((a, b) => a + b);
let leftSum = 0;
for (let i = 0; i < arr.length; ++i) {
sum -= arr[i];
if (leftSum === sum) {
return i;
}
leftSum += arr[i];
}
return -1;
}
- Output:
3, -1, 1, 0, 0
ES6 Functional
A composition of pure generic functions, returning all equilibrium indices.
(() => {
"use strict";
// ------------- ALL EQUILIBRIUM INDICES -------------
// equilibriumIndices :: [Int] -> [Int]
const equilibriumIndices = xs =>
zip(
scanl1(add)(xs)
)(
scanr1(add)(xs)
)
.reduceRight(
(a, xy, i) => xy[0] === xy[1] ? (
[i, ...a]
) : a, []
);
// ---------------------- TEST -----------------------
const main = () => [
[-7, 1, 5, 2, -4, 3, 0],
[2, 4, 6],
[2, 9, 2],
[1, -1, 1, -1, 1, -1, 1],
[1],
[]
].map(compose(
JSON.stringify,
equilibriumIndices
))
.join("\n");
// -> [[3, 6], [], [1], [0, 1, 2, 3, 4, 5, 6], [0], []]
// ---------------- GENERIC FUNCTIONS ----------------
// add (+) :: Num a => a -> a -> a
const add = a =>
// Curried addition.
b => a + b;
// compose (<<<) :: (b -> c) -> (a -> b) -> a -> c
const compose = (...fs) =>
// A function defined by the right-to-left
// composition of all the functions in fs.
fs.reduce(
(f, g) => x => f(g(x)),
x => x
);
// scanl :: (b -> a -> b) -> b -> [a] -> [b]
const scanl = f => startValue => xs =>
// The series of interim values arising
// from a catamorphism. Parallel to foldl.
xs.reduce((a, x) => {
const v = f(a[0])(x);
return [v, a[1].concat(v)];
}, [startValue, [startValue]])[1];
// scanl1 :: (a -> a -> a) -> [a] -> [a]
const scanl1 = f =>
// scanl1 is a variant of scanl that has no
// starting value argument.
xs => xs.length > 0 ? (
scanl(f)(
xs[0]
)(xs.slice(1))
) : [];
// scanr :: (a -> b -> b) -> b -> [a] -> [b]
const scanr = f =>
startValue => xs => xs.reduceRight(
(a, x) => {
const v = f(x)(a[0]);
return [v, [v].concat(a[1])];
}, [startValue, [startValue]]
)[1];
// scanr1 :: (a -> a -> a) -> [a] -> [a]
const scanr1 = f =>
// scanr1 is a variant of scanr that has no
// seed-value argument, and assumes that
// xs is not empty.
xs => xs.length > 0 ? (
scanr(f)(
xs.slice(-1)[0]
)(xs.slice(0, -1))
) : [];
// zip :: [a] -> [b] -> [(a, b)]
const zip = xs =>
// The paired members of xs and ys, up to
// the length of the shorter of the two lists.
ys => Array.from({
length: Math.min(xs.length, ys.length)
}, (_, i) => [xs[i], ys[i]]);
// MAIN ---
return main();
})();
- Output:
[3,6] [] [1] [0,1,2,3,4,5,6] [0] []
Also works with gojq and jq, the Go and Rust implementations of jq
`equilibrium_indices` is defined as a 0-arity filter that emits answers as a stream, as is idiomatic in jq.
# The index origin is 0 in jq.
def equilibrium_indices:
. as $in
| add as $add
| foreach range(0;length) as $i (
[0, 0, $add]; # [before, pivot, after]
$in[$i] as $x | [.[0]+.[1], $x, .[2] - $x];
if .[0] == .[2] then $i else empty end) ;Example 1:
[-7, 1, 5, 2, -4, 3, 0] | equilibrium_indices- Output:
$ jq -M -n -f equilibrium_indices.jq 3 6
Example 2:
def count(g): reduce g as $i (0; .+1);
# Create an array of length n with "init" elements:
def array(n;init): reduce range(0;n) as $i ([]; . + [0]);
count( array(10000;0) | equilibrium_indices )- Output:
$ jq -M -n -f equilibrium_indices.jq 10000
function equindex2pass(data::Array)
rst = Int[]
suml, sumr, ddelayed = 0, sum(data), 0
for (i, d) in enumerate(data)
suml += ddelayed
sumr -= d
ddelayed = d
if suml == sumr
push!(rst, i)
end
end
return rst
end
@show equindex2pass([1, -1, 1, -1, 1, -1, 1])
@show equindex2pass([1, 2, 2, 1])
@show equindex2pass([-7, 1, 5, 2, -4, 3, 0])
- Output:
equindex2pass([1, -1, 1, -1, 1, -1, 1]) = [1, 2, 3, 4, 5, 6, 7] equindex2pass([1, 2, 2, 1]) = Int64[] equindex2pass([-7, 1, 5, 2, -4, 3, 0]) = [4, 7]
f:{&{(+/y# x)=+/(y+1)_x}[x]'!#x}
f -7 1 5 2 -4 3 0
3 6
f 2 4 6
!0
f 2 9 2
,1
f 1 -1 1 -1 1 -1 1
0 1 2 3 4 5 6
// version 1.1
fun equilibriumIndices(a: IntArray): MutableList<Int> {
val ei = mutableListOf<Int>()
if (a.isEmpty()) return ei // empty list
val sumAll = a.sumOf { it }
var sumLeft = 0
var sumRight: Int
for (i in 0 until a.size) {
sumRight = sumAll - sumLeft - a[i]
if (sumLeft == sumRight) ei.add(i)
sumLeft += a[i]
}
return ei
}
fun main(args: Array<String>) {
val a = intArrayOf(-7, 1, 5, 2, -4, 3, 0)
val ei = equilibriumIndices(a)
when (ei.size) {
0 -> println("There are no equilibrium indices")
1 -> println("The only equilibrium index is : ${ei[0]}")
else -> println("The equilibrium indices are : ${ei.joinToString(", ")}")
}
}
- Output:
The equilibrium indices are : 3, 6
program equilibrium;
(* Equilibrium index *)
var
x1, x2, x3, x4: arrayof integer,
x1_eqs, x2_eqs, x3_eqs, x4_eqs: arrayof boolean;
unit write_array: procedure (nums: arrayof integer; nl: boolean);
var
i: integer;
begin
for i := lower(nums) to upper(nums)
do
write(nums(i): 1, " ")
od;
if nl then writeln fi
end write_array;
unit write_inds_of_true: procedure (bools: arrayof boolean; nl: boolean);
var
i: integer;
begin
for i := lower(bools) to upper(bools)
do
if bools(i) then write(i: 1, " ") fi;
od;
if nl then writeln fi
end write_inds_of_true;
unit get_inds: procedure (nums: arrayof integer; bools: arrayof boolean);
var
left_sum, right_sum: integer,
i: integer;
begin
left_sum, right_sum := 0;
for i := lower(nums) to upper(nums)
do
right_sum := right_sum + nums(i)
od;
for i := lower(nums) to upper(nums)
do
right_sum := right_sum - nums(i);
bools(i) := (left_sum = right_sum);
left_sum := left_sum + nums(i);
od
end get_inds;
begin
array x1 dim (0 : 6);
array x1_eqs dim (0 : 6);
x1(0) := -7; x1(1) := 1; x1(2) := 5; x1(3) := 2;
x1(4) := -4; x1(5) := 3; x1(6) := 0;
call get_inds(x1, x1_eqs);
array x2 dim (0 : 2);
array x2_eqs dim (0 : 2);
x2(0) := 2; x2(1) := 4; x2(2) := 6;
call get_inds(x2, x2_eqs);
array x3 dim (0 : 2);
array x3_eqs dim (0 : 2);
x3(0) := 2; x3(1) := 9; x3(2) := 2;
call get_inds(x3, x3_eqs);
array x4 dim (0 : 6);
array x4_eqs dim (0 : 6);
x4(0) := 1; x4(1) := -1; x4(2) := 1; x4(3) := -1;
x4(4) := 1; x4(5) := -1; x4(6) := 1;
call get_inds(x4, x4_eqs);
writeln("Results:");
writeln;
write("X1: ");
call write_array(x1, true);
write("Eqs: ");
call write_inds_of_true(x1_eqs, true);
writeln;
write("X2: ");
call write_array(x2, true);
write("Eqs: ");
call write_inds_of_true(x2_eqs, true);
writeln;
write("X3: ");
call write_array(x3, true);
write("Eqs: ");
call write_inds_of_true(x3_eqs, true);
writeln;
write("X4: ");
call write_array(x4, true);
write("Eqs: ");
call write_inds_of_true(x4_eqs, true);
end equilibrium.- Output:
Results: X1: -7 1 5 2 -4 3 0 Eqs: 3 6 X2: 2 4 6 Eqs: X3: 2 9 2 Eqs: 1 X4: 1 -1 1 -1 1 -1 1 Eqs: 0 1 2 3 4 5 6
to equilibrium.iter :i :before :after :tail :ret
if equal? :before :after [make "ret lput :i :ret]
if empty? butfirst :tail [output :ret]
output equilibrium.iter :i+1 (:before+first :tail) (:after-first butfirst :tail) (butfirst :tail) :ret
end
to equilibrium.index :list
output equilibrium.iter 1 0 (apply "sum butfirst :list) :list []
end
show equilibrium_index [-7 1 5 2 -4 3 0] ; [4 7]function array_sum(t)
assert(type(t) == "table", "t must be a table!")
local sum = 0
for i=1, #t do sum = sum + t[i] end
return sum
end
function equilibrium_index(t)
assert(type(t) == "table", "t must be a table!")
local left, right, ret = 0, array_sum(t), -1
for i,j in pairs(t) do
right = right - j
if left == right then
ret = i
break
end
left = left + j
end
return ret
end
print(equilibrium_index({-7, 1, 5, 2, -4, 3, 0}))
Module Equilibrium_index{
Function equilibriumIndices(a as tuple) {
flush ' empty stack of values
sum=a#sum()
b=each(a)
double sumleft, sumright
while b
sumright=sum-sumleft-array(b)
if sumleft=sumright then
data b^ ' put index to end of stack (used as FIFO)
end if
sumleft+=array(b)
end while
=array([])
}
a=(-7, 1, 5, 2, -4, 3, 0)
b=equilibriumIndices(a)#str$()
Print "["+b+"]"
Print "["+equilibriumIndices((2, 4, 6))#str$()+"]"
Print "["+equilibriumIndices((2, 9, 2))#str$()+"]"
Print "["+equilibriumIndices((1, -1, 1, -1, 1, -1, 1))#str$()+"]"
}
Equilibrium_index
- Output:
[3 6] [] [1] [0 1 2 3 4 5 6]
Mathematica indexes are 1-based so the output of this program will be shifted up by one compared to solutions in languages with 0-based arrays.
equilibriumIndex[data_]:=Reap[
Do[If[Total[data[[;; n - 1]]] == Total[data[[n + 1 ;;]]],Sow[n]],
{n, Length[data]}]][[2, 1]]
- Usage:
equilibriumIndex[{-7 , 1, 5 , 2, -4 , 3, 0}]
{4, 7}
MATLAB arrays are 1-based so the output of this program will be shifted up by one compared to solutions in languages with 0-based arrays.
function indicies = equilibriumIndex(list)
indicies = [];
for i = (1:numel(list))
if ( sum(-list(1:i)) == sum(-list(i:end)) )
indicies = [indicies i];
end
end
end
- Output:
>> equilibriumIndex([-7 1 5 2 -4 3 0])
ans =
4 7
MODULE Equilibrium;
(* Equilibrium index *)
FROM STextIO IMPORT
WriteLn, WriteString;
FROM SWholeIO IMPORT
WriteInt;
TYPE
(* These types are for easier initialization in the main program.*)
TData7Items = ARRAY [0 .. 6] OF INTEGER;
TData3Items = ARRAY [0 .. 2] OF INTEGER;
VAR
X1: TData7Items;
X1Eqs: ARRAY [0 .. 6] OF BOOLEAN;
X2: TData3Items;
X2Eqs: ARRAY [0 .. 2] OF BOOLEAN;
X3: TData3Items;
X3Eqs: ARRAY [0 .. 2] OF BOOLEAN;
X4: TData7Items;
X4Eqs: ARRAY [0 .. 6] OF BOOLEAN;
PROCEDURE WriteArray(Nums: ARRAY OF INTEGER; NL: BOOLEAN);
VAR
I: CARDINAL;
BEGIN
FOR I := 0 TO HIGH(Nums) DO
WriteInt(Nums[I], 1);
WriteString(" ")
END;
IF NL THEN
WriteLn
END
END WriteArray;
PROCEDURE WriteIndicesOfTrue(Bools: ARRAY OF BOOLEAN; NL: BOOLEAN);
VAR
I: CARDINAL;
BEGIN
FOR I := 0 TO HIGH(Bools) DO
IF Bools[I] THEN
WriteInt(I, 1);
WriteString(" ")
END;
END;
IF NL THEN
WriteLn
END
END WriteIndicesOfTrue;
PROCEDURE GetIndices(Nums: ARRAY OF INTEGER; VAR OUT Result: ARRAY OF BOOLEAN);
VAR
LeftSum, RightSum: INTEGER;
I: CARDINAL;
BEGIN
LeftSum := 0;
RightSum := 0;
FOR I := 0 TO HIGH(Nums) DO
RightSum := RightSum + Nums[I]
END;
FOR I := 0 TO HIGH(Nums) DO
RightSum := RightSum - Nums[I];
Result[I] := (LeftSum = RightSum);
LeftSum := LeftSum + Nums[I]
END;
END GetIndices;
BEGIN
X1 := TData7Items{-7, 1, 5, 2, -4, 3, 0};
GetIndices(X1, X1Eqs);
X2 := TData3Items{2, 4, 6};
GetIndices(X2, X2Eqs);
X3 := TData3Items{2, 9, 2};
GetIndices(X3, X3Eqs);
X4 := TData7Items{1, -1, 1, -1, 1 ,-1, 1};
GetIndices(X4, X4Eqs);
WriteString("Results:");
WriteLn;
WriteLn;
WriteString("X1: ");
WriteArray(X1, TRUE);
WriteString("Eqs: ");
WriteIndicesOfTrue(X1Eqs, TRUE);
WriteLn;
WriteString("X2: ");
WriteArray(X2, TRUE);
WriteString("Eqs: ");
WriteIndicesOfTrue(X2Eqs, TRUE);
WriteLn;
WriteString("X3: ");
WriteArray(X3, TRUE);
WriteString("Eqs: ");
WriteIndicesOfTrue(X3Eqs, TRUE);
WriteLn;
WriteString("X4: ");
WriteArray(X4, TRUE);
WriteString("Eqs: ");
WriteIndicesOfTrue(X4Eqs, TRUE);
END Equilibrium.
- Output:
Results: X1: -7 1 5 2 -4 3 0 Eqs: 3 6 X2: 2 4 6 Eqs: X3: 2 9 2 Eqs: 1 X4: 1 -1 1 -1 1 -1 1 Eqs: 0 1 2 3 4 5 6
/* NetRexx */
options replace format comments java crossref symbols nobinary
numeric digits 20
runSample(arg)
return
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
-- @see http://www.geeksforgeeks.org/equilibrium-index-of-an-array/
method equilibriumIndex(sequence) private static
es = ''
loop ix = 1 to sequence.words()
sum = 0
loop jx = 1 to sequence.words()
if jx < ix then sum = sum + sequence.word(jx)
if jx > ix then sum = sum - sequence.word(jx)
end jx
if sum = 0 then es = es ix
end ix
return es
-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method runSample(arg) private static
-- Note: A Rexx object based list of "words" starts at index 1
sequences = [ -
'-7 1 5 2 -4 3 0', - -- 4 7
' 2 4 6' , - -- (no equilibrium point)
' 0 2 4 0 6 0' , - -- 4
' 2 9 2' , - -- 2
' 1 -1 1 -1 1 -1 1' - -- 1 2 3 4 5 6 7
]
loop sequence over sequences
say 'For sequence "'sequence.space(1, ',')'" the equilibrium indices are: \-'
say '"'equilibriumIndex(sequence).space(1, ',')'"'
end sequence
return
- Output:
For sequence "-7,1,5,2,-4,3,0" the equilibrium indices are: "4,7" For sequence "2,4,6" the equilibrium indices are: "" For sequence "0,2,4,0,6,0" the equilibrium indices are: "4" For sequence "2,9,2" the equilibrium indices are: "2" For sequence "1,-1,1,-1,1,-1,1" the equilibrium indices are: "1,2,3,4,5,6,7"
import math, sequtils, strutils
iterator eqindex(data: openArray[int]): int =
var suml, ddelayed = 0
var sumr = sum(data)
for i,d in data:
suml += ddelayed
sumr -= d
ddelayed = d
if suml == sumr:
yield i
const d = @[@[-7, 1, 5, 2, -4, 3, 0],
@[2, 4, 6],
@[2, 9, 2],
@[1, -1, 1, -1, 1, -1, 1]]
for data in d:
echo "d = [", data.join(", "), ']'
echo "eqIndex(d) -> [", toSeq(eqindex(data)).join(", "), ']'
- Output:
d = [-7, 1, 5, 2, -4, 3, 0] eqIndex(d) -> [3, 6] d = [2, 4, 6] eqIndex(d) -> [] d = [2, 9, 2] eqIndex(d) -> [1] d = [1, -1, 1, -1, 1, -1, 1] eqIndex(d) -> [0, 1, 2, 3, 4, 5, 6]
class Rosetta {
function : Main(args : String[]) ~ Nil {
sequence := [-7, 1, 5, 2, -4, 3, 0];
EqulibriumIndices(sequence);
}
function : EqulibriumIndices(sequence : Int[]) ~ Nil {
# find total sum
totalSum := 0;
each(i : sequence) {
totalSum += sequence[i];
};
# compare running sum to remaining sum to find equlibrium indices
runningSum := 0;
each(i : sequence) {
n := sequence[i];
if (totalSum - runningSum - n = runningSum) {
i->PrintLine();
};
runningSum += n;
};
}
}Output:
3 6
let lst = [ -7; 1; 5; 2; -4; 3; 0 ]
let sum = List.fold_left ( + ) 0 lst
let () =
let rec aux acc i left right = function
| x::xs ->
let right = right - x in
let acc = if left = right then i::acc else acc in
aux acc (succ i) (left + x) right xs
| [] -> List.rev acc
in
let res = aux [] 0 0 sum lst in
print_string "Results:";
List.iter (Printf.printf " %d") res;
print_newline ()
Oforth collections are 1-based
: equilibrium(l)
| ls rs i e |
0 ->ls
l sum ->rs
ListBuffer new l size loop: i [
l at(i) ->e
rs e - dup ->rs ls == ifTrue: [ i over add ]
ls e + ->ls
] ;- Output:
>equilibrium([-7, 1, 5, 2, -4, 3, 0]) println [4, 7]
This uses 1-based vectors instead of 0-based arrays; subtract 1 from each index if you prefer the other style.
equilib(v)={
my(a=sum(i=2,#v,v[i]),b=0,u=[]);
for(i=1,#v-1,
if(a==b, u=concat(u,i));
b+=v[i];
a-=v[i+1]
);
if(b,u,concat(u,#v))
};Program EquilibriumIndexDemo(output);
{$IFDEF FPC}{$Mode delphi}{$ENDIF}
function ArraySum(list: array of integer; first, last: integer): integer;
var
i: integer;
begin
Result := 0;
for i := first to last do // not taken if first > last
Result := Result + list[i];
end;
procedure EquilibriumIndex(list: array of integer; offset: integer);
var
i: integer;
begin
for i := low(list) to high(list) do
if ArraySum(list, low(list), i-1) = ArraySum(list, i+1, high(list)) then
write(offset + i:3);
end;
var
{** The base index of the array is fully taken care off and can be any number. **}
numbers: array [1..7] of integer = (-7, 1, 5, 2, -4, 3, 0);
i: integer;
begin
write('List of numbers: ');
for i := low(numbers) to high(numbers) do
write(numbers[i]:3);
writeln;
write('Equilibirum indices: ');
EquilibriumIndex(numbers, low(numbers));
writeln;
end.
- Output:
:> ./EquilibriumIndex List of numbers: -7 1 5 2 -4 3 0 Equilibirum indices: 4 7
alternative
slightly modified.Calculating the sum only once.Using a zero-based array type.Data type could be any type of signed integer or float. But beware, that during building the sum, the limits of the data type mustn't be violated.
Program EquilibriumIndexDemo(output);
{$IFDEF FPC}{$Mode delphi}{$ENDIF}
type
tEquiData = shortInt;//Int64;extended ,double
tnumList = array of tEquiData;
tresList = array of LongInt;
const
cNumbers: array [11..17] of tEquiData = (-7, 1, 5, 2, -4, 3, 0);
function ArraySum(const list: tnumList):tEquiData;
var
i: integer;
begin
result := 0;
for i := Low(list) to High(list) do
result := result+list[i];
end;
procedure EquilibriumIndex(const list:tnumList;
var indices:tresList);
var
pC : ^tEquiData;
LeftSum,
RightSum : tEquiData;
i,idx,HiList: integer;
begin
HiList := High(List);
RightSum :=ArraySum(list);
setlength(indices,10);
idx := 0;
i := -Hilist;
pC := @List[0];
LeftSum:= 0;
repeat
Rightsum:= RightSum-pC^;
IF LeftSum = RightSum then
Begin
indices[idx] := Hilist+i;
inc(idx);
IF idx > high(indices) then
setlength(indices, idx+10);
end;
inc(i);
leftSum := leftsum+pC^;
inc(pC);
until i>=0;
leftSum := leftsum+pC^;
IF LeftSum = RightSum then
Begin
indices[idx] := Hilist+i;
inc(idx);
end;
setlength(indices,idx);
end;
procedure TestRun(const numbers:tnumList);
var
indices : tresList;
i: integer;
Begin
write('List of numbers: ');
for i := low(numbers) to high(numbers) do
write(numbers[i]:3);
writeln;
EquilibriumIndex(numbers,indices);
write('Equilibirum indices: ');
EquilibriumIndex(numbers,indices);
for i := low(indices) to high(indices) do
write(indices[i]:3);
writeln;
writeln;
end;
var
numbers: tnumList;
I: integer;
begin
setlength(numbers,High(cNumbers)-Low(cNumbers)+1);
move(cNumbers[Low(cNumbers)],numbers[0],sizeof(cnumbers));
TestRun(numbers);
for i := low(numbers) to high(numbers) do
numbers[i]:= 0;
TestRun(numbers);
end.
- Output:
List of numbers: -7 1 5 2 -4 3 0Equilibirum indices: 3 6
List of numbers: 0 0 0 0 0 0 0
Equilibirum indices: 0 1 2 3 4 5 6
##
function eqindex2Pass(data: array of integer): sequence of integer;
begin
var (suml, sumr, ddelayed) := (0, data.sum, 0);
foreach var d in data index i do
begin
suml += ddelayed;
sumr -= d;
ddelayed := d;
if suml = sumr then
yield i
end;
end;
var d := ||-7, 1, 5, 2, -4, 3, 0|, |2, 4, 6|, |2, 9, 2|, |1, -1, 1, -1, 1, -1, 1||;
foreach var data in d do
println('d =', data, '->', eqindex2pass(data));
- Output:
d = [-7,1,5,2,-4,3,0] -> [3,6] d = [2,4,6] -> [] d = [2,9,2] -> [1] d = [1,-1,1,-1,1,-1,1] -> [0,1,2,3,4,5,6]
sub eq_index {
my ( $i, $sum, %sums ) = ( 0, 0 );
for (@_) {
push @{ $sums{ $sum * 2 + $_ } }, $i++;
$sum += $_;
}
return join ' ', @{ $sums{$sum} || [] }, "\n";
}
print eq_index qw( -7 1 5 2 -4 3 0 ); # 3 6
print eq_index qw( 2 4 6 ); # (no eq point)
print eq_index qw( 2 9 2 ); # 1
print eq_index qw( 1 -1 1 -1 1 -1 1 ); # 0 1 2 3 4 5 6
with javascript_semantics
function equilibrium(sequence s)
atom lower_sum = 0,
higher_sum = sum(s)
sequence res = {}
for i,si in s do
higher_sum -= si
if lower_sum=higher_sum then
res &= i
end if
lower_sum += si
end for
return res
end function
constant tests = {{-7,1,5,2,-4,3,0},
{2,4,6},{2,9,2},
{1,-1,1,-1,1,-1,1},
{1},{}}
printf(1,"The equilibrium indices for the following sequences are:\n")
for t in tests do
printf(1,"%24v -> %v\n", {t,equilibrium(t)})
end for
- Output:
(Remember that indices are 1-based in Phix)
The equilibrium indices for the following sequences are:
{-7,1,5,2,-4,3,0} -> {4,7}
{2,4,6} -> {}
{2,9,2} -> {2}
{1,-1,1,-1,1,-1,1} -> {1,2,3,4,5,6,7}
{1} -> {1}
{} -> {}
<?php
$arr = array(-7, 1, 5, 2, -4, 3, 0);
function getEquilibriums($arr) {
$right = array_sum($arr);
$left = 0;
$equilibriums = array();
foreach($arr as $key => $value){
$right -= $value;
if($left == $right) $equilibriums[] = $key;
$left += $value;
}
return $equilibriums;
}
echo "# results:\n";
foreach (getEquilibriums($arr) as $r) echo "$r, ";
?>
- Output:
# results: 3, 6,
Note: Picat is 1-based.
Prolog-style
equilibrium_index1(A, Ix) =>
append(Front, [_|Back], A),
sum(Front) = sum(Back),
Ix = length(Front)+1. % give 1 based indexLoop approach
equilibrium_index2(A, Ix) =>
Len = A.length,
Ix1 = [],
TotalSum = sum(A),
RunningSum = 0,
foreach(I in 1..Len)
AI = A[I],
if TotalSum - RunningSum - AI == RunningSum then
Ix1 := Ix1 ++ [I]
end,
RunningSum := RunningSum + AI
end,
Ix = Ix1.Test the approaches.
go =>
As = [
[-7, 1, 5, 2, -4, 3, 0], % 4 7
[ 2, 4, 6], % (no equilibrium point)
[ 0, 2, 4, 0, 6, 0], % 4
[ 2, 9, 2], % 2
[ 1, -1, 1, -1, 1, -1, 1] % 1 2 3 4 5 6 7
],
foreach(A in As)
println(a=A),
All1 = findall(Ix1, equilibrium_index1(A,Ix1)),
println(all1=All1),
equilibrium_index2(A,All2),
println(all2=All2),
nl
end,
% A larger random instance
print("A larger random instance:"),
_ = random2(),
N = 5001,
Random = [random(-10,10) : _ in 1..N],
% println(Random),
time(R1 = findall(IxR1, equilibrium_index1(Random,IxR1))),
println(r1=R1),
time(equilibrium_index2(Random,R2)),
println(r2=R2),
nl.- Output:
a = [-7,1,5,2,-4,3,0] all1 = [4,7] all2 = [4,7] a = [2,4,6] all1 = [] all2 = [] a = [0,2,4,0,6,0] all1 = [4] all2 = [4] a = [2,9,2] all1 = [2] all2 = [2] a = [1,-1,1,-1,1,-1,1] all1 = [1,2,3,4,5,6,7] all2 = [1,2,3,4,5,6,7] A larger random instance: CPU time 0.113 seconds. r1 = [115,372,4082,4254,4258,4261] CPU time 0.019 seconds. r2 = [115,372,4082,4254,4258,4261]
(de equilibria (Lst)
(make
(let Sum 0
(for ((I . L) Lst L (cdr L))
(and (= Sum (sum prog (cdr L))) (link I))
(inc 'Sum (car L)) ) ) ) )- Output:
: (equilibria (-7 1 5 2 -4 3 0)) -> (4 7) : (equilibria (make (do 10000 (link (rand -10 10))))) -> (4091 6174 6198 7104 7112 7754)
local fmt = require "fmt"
local function equilibrium(a)
local len = #a
local equi = {}
if len == 0 then return equi end -- sequence has no indices at all
local rsum = a:reduce(|acc, x| -> acc + x)
local lsum = 0
for i = 1, len do
rsum -= a[i]
if rsum == lsum then equi:insert(i) end
lsum += a[i]
end
return equi
end
local tests = {
{-7, 1, 5, 2, -4, 3, 0},
{2, 4, 6},
{2, 9, 2},
{1, -1, 1, -1, 1, -1, 1},
{1},
{}
}
print("The equilibrium indices for the following sequences are:\n")
for tests as test do
fmt.print("%24s -> %s", fmt.swrite(test), fmt.swrite(equilibrium(test)))
end
- Output:
Note that array indices are normally 1-based in Pluto.
The equilibrium indices for the following sequences are:
{-7, 1, 5, 2, -4, 3, 0} -> {4, 7}
{2, 4, 6} -> {}
{2, 9, 2} -> {2}
{1, -1, 1, -1, 1, -1, 1} -> {1, 2, 3, 4, 5, 6, 7}
{1} -> {1}
{} -> {}
In real life in PowerShell, one would likely leverage pipelines, ForEach-Object, Where-Object, and Measure-Object for tasks such as this. Normally in PowerShell, speed is an important, but not primary consideration, and the advantages of pipelines tend to outweigh the overhead incurred. However, for this particular task, keeping in mind that “the sequence may be very long,” this code was optimized primarly for speed.
function Get-EquilibriumIndex ( $Sequence )
{
$Indexes = 0..($Sequence.Count - 1)
$EqulibriumIndex = @()
ForEach ( $TestIndex in $Indexes )
{
$Left = 0
$Right = 0
ForEach ( $Index in $Indexes )
{
If ( $Index -lt $TestIndex ) { $Left += $Sequence[$Index] }
ElseIf ( $Index -gt $TestIndex ) { $Right += $Sequence[$Index] }
}
If ( $Left -eq $Right )
{
$EqulibriumIndex += $TestIndex
}
}
return $EqulibriumIndex
}
Get-EquilibriumIndex -7, 1, 5, 2, -4, 3, 0
- Output:
3 6
equilibrium_index(List, Index) :-
append(Front, [_|Back], List),
sumlist(Front, Sum),
sumlist(Back, Sum),
length(Front, Len),
Index is Len.
Example:
?- equilibrium_index([-7, 1, 5, 2, -4, 3, 0], Index).
Index = 3 ;
Index = 6 ;
false.
Two Pass
Uses an initial summation of the whole list then visits each item of the list adding it to the left-hand sum (after a delay); and subtracting the item from the right-hand sum. I think it should be quicker than algorithms that scan the list creating left and right sums for each index as it does ~2N add/subtractions rather than n*n.
def eqindex2Pass(data):
"Two pass"
suml, sumr, ddelayed = 0, sum(data), 0
for i, d in enumerate(data):
suml += ddelayed
sumr -= d
ddelayed = d
if suml == sumr:
yield i
Multi Pass
This is the version that does more summations, but may be faster for some sizes of input as the sum function is implemented in C internally:
def eqindexMultiPass(data):
"Multi pass"
for i in range(len(data)):
suml, sumr = sum(data[:i]), sum(data[i+1:])
if suml == sumr:
yield i
Shorter alternative:
def eqindexMultiPass(s):
return [i for i in xrange(len(s)) if sum(s[:i]) == sum(s[i+1:])]
print eqindexMultiPass([-7, 1, 5, 2, -4, 3, 0])
One Pass
This routine would need careful evaluation against the two-pass solution above as, although it only runs through the data once, it may create a dict that is as long as the input data in its worst case of an input of say a simple 1, 2, 3, ... counting sequence.
from collections import defaultdict
def eqindex1Pass(data):
"One pass"
l, h = 0, defaultdict(list)
for i, c in enumerate(data):
l += c
h[l * 2 - c].append(i)
return h[l]
Tests
f = (eqindex2Pass, eqindexMultiPass, eqindex1Pass)
d = ([-7, 1, 5, 2, -4, 3, 0],
[2, 4, 6],
[2, 9, 2],
[1, -1, 1, -1, 1, -1, 1])
for data in d:
print("d = %r" % data)
for func in f:
print(" %16s(d) -> %r" % (func.__name__, list(func(data))))
- Sample output:
d = [-7, 1, 5, 2, -4, 3, 0]
eqindex2Pass(d) -> [3, 6]
eqindexMultiPass(d) -> [3, 6]
eqindex1Pass(d) -> [3, 6]
d = [2, 4, 6]
eqindex2Pass(d) -> []
eqindexMultiPass(d) -> []
eqindex1Pass(d) -> []
d = [2, 9, 2]
eqindex2Pass(d) -> [1]
eqindexMultiPass(d) -> [1]
eqindex1Pass(d) -> [1]
d = [1, -1, 1, -1, 1, -1, 1]
eqindex2Pass(d) -> [0, 1, 2, 3, 4, 5, 6]
eqindexMultiPass(d) -> [0, 1, 2, 3, 4, 5, 6]
eqindex1Pass(d) -> [0, 1, 2, 3, 4, 5, 6]
In terms of itertools.accumulate
The left scan is efficiently derived by the accumulate function in the itertools module.
The right scan can be derived from the left as a map or equivalent list comprehension:
"""Equilibrium index"""
from itertools import (accumulate)
# equilibriumIndices :: [Num] -> [Int]
def equilibriumIndices(xs):
'''List indices at which the sum of values to the left
equals the sum of values to the right.
'''
def go(xs):
'''Left scan from accumulate,
right scan derived from left
'''
ls = list(accumulate(xs))
n = ls[-1]
return [
i for (i, (x, y)) in enumerate(zip(
ls,
[n] + [n - x for x in ls[0:-1]]
)) if x == y
]
return go(xs) if xs else []
# ------------------------- TEST -------------------------
# main :: IO ()
def main():
'''Tabulated test results'''
print(
tabulated('Equilibrium indices:\n')(
equilibriumIndices
)([
[-7, 1, 5, 2, -4, 3, 0],
[2, 4, 6],
[2, 9, 2],
[1, -1, 1, -1, 1, -1, 1],
[1],
[]
])
)
# ----------------------- GENERIC ------------------------
# tabulated :: String -> (a -> b) -> [a] -> String
def tabulated(s):
'''heading -> function -> input List
-> tabulated output string
'''
def go(f):
def width(x):
return len(str(x))
def cols(xs):
w = width(max(xs, key=width))
return s + '\n' + '\n'.join([
str(x).rjust(w, ' ') + ' -> ' + str(f(x))
for x in xs
])
return cols
return go
if __name__ == '__main__':
main()
- Output:
Equilibrium indices:
[-7, 1, 5, 2, -4, 3, 0] -> [3, 6]
[2, 4, 6] -> []
[2, 9, 2] -> [1]
[1, -1, 1, -1, 1, -1, 1] -> [0, 1, 2, 3, 4, 5, 6]
[1] -> [0]
[] -> []
[ dip [ [] [] 0 ]
witheach
[ + dup dip join ]
over [] swap
witheach
[ dip over - join ]
join
-1 split drop
witheach
[ over i^ peek = if
[ dip [ i^ join ] ] ]
drop ] is equilibria ( [ --> [ )
' [ -7 1 5 2 -4 3 0 ] equilibria echo- Output:
[ 3 6 ]
Indices in R start at 1.
is_equilibrium <- function(v, n){
lower <- sum(v[seq_along(v)<n])
higher <- sum(v[seq_along(v)>n])
lower==higher
}
all_equilibriums <- function(v) Filter(function(n) is_equilibrium(v, n), seq_along(v))
all_equilibriums(c(-7,1,5,2,-4,3,0))
- Output:
[1] 4 7
#lang racket
(define (subsums xs)
(for/fold ([sums '()] [sum 0]) ([x xs])
(values (cons (+ x sum) sums)
(+ x sum))))
(define (equivilibrium xs)
(define-values (sums total) (subsums xs))
(for/list ([sum (reverse sums)]
[x xs]
[i (in-naturals)]
#:when (= (- sum x) (- total sum)))
i))
(equivilibrium '(-7 1 5 2 -4 3 0))
- Output:
'(3 6)
(formerly Perl 6)
sub equilibrium_index(@list) {
my ($left,$right) = 0, [+] @list;
gather for @list.kv -> $i, $x {
$right -= $x;
take $i if $left == $right;
$left += $x;
}
}
my @list = -7, 1, 5, 2, -4, 3, 0;
.say for equilibrium_index(@list).grep(/\d/);
And here's an FP solution that manages to remain O(n):
sub equilibrium_index(@list) {
my @a = [\+] @list;
my @b = reverse [\+] reverse @list;
^@list Zxx (@a »==« @b);
}
The [\+] is a reduction that returns a list of partial results. The »==« is a vectorized equality comparison; it returns a vector of true and false. The Zxx is a zip with the list replication operator, so we return only the elements of the left list where the right list is true (which is taken to mean 1 here). And the ^@list is just shorthand for 0 ..^ @list. We could just as easily have used @list.keys there.
Single-pass solution
The task can be restated in a way that removes the "right side" from the calculation.
C is the current element,
L is the sum of elements left of C,
R is the sum of elements right of C,
S is the sum of the entire list.
By definition, L + C + R == S for any choice of C, and L == R for any C that is an equilibrium point.
Therefore (by substituting L for R), L + C + L == S at all equilibrium points.
Restated, 2L + C == S.
# Original example, with expanded calculations:
0 1 2 3 4 5 6 # Index
-7 1 5 2 -4 3 0 # C (Value at index)
0 -7 -6 -1 1 -3 0 # L (Sum of left)
-7 -13 -7 0 -2 -3 0 # 2L+C
If we build a hash as we walk the list, with 2L+C as hash keys, and arrays of C-indexes as hash values, we get:
{
-7 => [ 0, 2 ],
-13 => [ 1 ],
0 => [ 3, 6 ],
-2 => [ 4 ],
-3 => [ 5 ],
}
After we have finished walking the list, we will have the sum (S), which we look up in the hash. Here S=0, so the equilibrium points are 3 and 6.
Note: In the code below, it is more convenient to calculate 2L+C *after* L has already been incremented by C; the calculation is simply 2L-C, because each L has an extra C in it. 2(L-C)+C == 2L-C.
sub eq_index ( *@list ) {
my $sum = 0;
my %h = @list.keys.classify: {
$sum += @list[$_];
$sum * 2 - @list[$_];
};
return %h{$sum} // [];
}
say eq_index < -7 1 5 2 -4 3 0 >; # 3 6
say eq_index < 2 4 6 >; # (no eq point)
say eq_index < 2 9 2 >; # 1
say eq_index < 1 -1 1 -1 1 -1 1 >; # 0 1 2 3 4 5 6
The .classify method creates a hash, with its code block's return value as key. Each hash value is an Array of all the inputs that returned that key.
We could have used .pairs instead of .keys to save the cost of @list lookups, but that would change each %h value to an Array of Pairs, which would complicate the return line.
Red code bellow is fully compatible with Rebol.
Red []
eqindex: func [a [block!]] [
collect [
repeat ind length? a [ if (sum skip a ind) = sum copy/part a ind - 1 [ keep ind ] ]
]
]
prin "(1 based) equ indices are: "
probe eqindex [-7 1 5 2 -4 3 0]
- Output:
(1 based) equ indices are: [4 7]
let arr = [-7, 1, 5, 2, -4, 3, 0]
let sum = Js.Array2.reduce(arr, \"+", 0)
let len = Js.Array.length(arr)
let rec aux = (acc, i, left, right) => {
if (i >= len) { acc } else {
let x = arr[i]
let right = right - x
if (left == right) {
let _ = Js.Array2.push(acc, i)
}
aux(acc, i+1, (left + x), right)
}
}
let res = aux([], 0, 0, sum)
Js.log("Results:")
Js.Array2.forEach(res, Js.log)version 1
This REXX version utilizes a zero-based stemmed array to mimic the illustrative example in this Rosetta Code task's
prologue, which uses a zero-based index.
/*REXX program calculates and displays the equilibrium index for a numeric array (list).*/
parse arg x /*obtain the optional arguments from CL*/
if x='' then x= copies(" 7 -7", 50) 7 /*Not specified? Then use the default.*/
say ' array list: ' space(x) /*echo the array list to the terminal. */
#= words(x) /*the number of elements in the X list.*/
do j=0 for #; @.j= word(x, j+1) /*zero─start is for zero─based array. */
end /*j*/ /* [↑] assign @.0 @.1 @.3 ··· */
say /* ··· and also display a blank line. */
answer= equilibriumIDX(); w= words(answer) /*calculate the equilibrium index. */
say 'equilibrium' word("(none) index: indices:", 1 + (w>0) + (w>1)) answer
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
equilibriumIDX: $=; do i=0 for #; sum= 0
do k=0 for #; sum= sum + @.k * sign(k-i); end /*k*/
if sum==0 then $= $ i
end /*i*/ /* [↑] Zero? Found an equilibrium index*/
return $ /*return equilibrium list (may be null)*/
- output when using the input of: -7 1 5 2 -4 3 0
array list: -7 1 5 2 -4 3 0 equilibrium indices: 3 6
- output when using the input of: 2 9 2
array list: 2 9 2 equilibrium index: 1
- output when using the input of: 5 4 4 5
array list: 5 4 4 5 equilibrium (none)
- output when using the default input
array list: 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 -7 7 equilibrium indices: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100
version 2
/* REXX ---------------------------------------------------------------
* 30.06.2014 Walter Pachl
*--------------------------------------------------------------------*/
parse arg l
say ' array list:' strip(l)
x.=0
Do i=1 To words(l)
x.i=word(l,i)
End
n=i-1
ans=strip(equilibriumIndices())
n=words(ans)
Select
When n=0 Then Say 'There''s no equilibrium index'
When n=1 Then Say 'equilibrium index :' ans
Otherwise Say 'equilibrium indices:' ans
End
Say '---'
exit
equilibriumIndices: procedure expose x. n
sum.=0
sum=0
eil=''
Do i=1 To n
sum=sum+x.i
sum.i=sum
End
Do i=1 To n
im1=i-1
If sum.im1=(sum.n-x.i)/2 Then
eil=eil im1
End
Return eil
output
array list: -7 1 5 2 -4 3 0
equilibrium indices: 3 6
---
array list: 2 9 2
equilibrium index : 1
---
array list: 1 -1 1 -1 1 -1 1
equilibrium indices: 0 1 2 3 4 5 6
---
array list: 1 1
There's no equilibrium index
---
list = [-7, 1, 5, 2, -4, 3, 0]
see "equilibrium indices are : " + equilibrium(list) + nl
func equilibrium l
r = 0 s = 0 e = ""
for n = 1 to len(l)
s += l[n]
next
for i = 1 to len(l)
if r = s - r - l[i] e += string(i-1) + "," ok
r += l[i]
next
e = left(e,len(e)-1)
return eOutput:
equilibrium indices are : 3,6
| RPL code | Comment |
|---|---|
≪
0 SWAP + → seq
≪ { } 0 seq ∑LIST
2 seq SIZE FOR j
seq j GET - SWAP seq j 1 - GET + SWAP
IF DUP2 == THEN ROT j 2 - + ROT ROT END
NEXT DROP2
≫ ≫ ‘EQIDX’ STO
|
EQIDX ( { A0..An } -- { equilibrium index } )
add zero at list head to avoid GET error at first loop
left = 0 ; right = A0+A1+...An
loop from j=2 to length(seq) e.g. A0 to An
right -= seq[j] ; left += A[j-1]
if left = right then append j-2 to index list
drop left and right
return list
|
{ -7 1 5 2 -4 3 0 } EQIDX
- Output:
1: { 3 6 }
- Functional Style
def eq_indices(list)
list.each_index.select do |i|
list[0...i].sum == list[i+1..-1].sum
end
end
- Tail Recursion
- This one would be good if Ruby did tail-call optimization (TCO).
- MRI does not do TCO; so this function fails with a long list (by overflowing the call stack).
def eq_indices(list)
result = []
list.empty? and return result
final = list.size - 1
helper = lambda do |left, current, right, index|
left == right and result << index # Push index to result?
index == final and return # Terminate recursion?
new = list[index + 1]
helper.call(left + current, new, right - new, index + 1)
end
helper.call 0, list.first, list.drop(1).sum, 0
result
end
- Imperative Style (faster)
def eq_indices(list)
left, right = 0, list.sum
equilibrium_indices = []
list.each_with_index do |val, i|
right -= val
equilibrium_indices << i if right == left
left += val
end
equilibrium_indices
end
- Test
indices = [
[-7, 1, 5, 2,-4, 3, 0],
[2, 4, 6],
[2, 9, 2],
[1,-1, 1,-1, 1,-1, 1]
]
indices.each do |x|
puts "%p => %p" % [x, eq_indices(x)]
end
- Output:
[-7, 1, 5, 2, -4, 3, 0] => [3, 6] [2, 4, 6] => [] [2, 9, 2] => [1] [1, -1, 1, -1, 1, -1, 1] => [0, 1, 2, 3, 4, 5, 6]
extern crate num;
use num::traits::Zero;
fn equilibrium_indices(v: &[i32]) -> Vec<usize> {
let mut right = v.iter().sum();
let mut left = i32::zero();
v.iter().enumerate().fold(vec![], |mut out, (i, &el)| {
right -= el;
if left == right {
out.push(i);
}
left += el;
out
})
}
fn main() {
let v = [-7i32, 1, 5, 2, -4, 3, 0];
let indices = equilibrium_indices(&v);
println!("Equilibrium indices for {:?} are: {:?}", v, indices);
}
- Output:
Equilibrium indices for [-7, 1, 5, 2, -4, 3, 0] are: [3, 6]
def getEquilibriumIndex(A: Array[Int]): Int = {
val bigA: Array[BigInt] = A.map(BigInt(_))
val partialSums: Array[BigInt] = bigA.scanLeft(BigInt(0))(_+_).tail
def lSum(i: Int): BigInt = if (i == 0) 0 else partialSums(i - 1)
def rSum(i: Int): BigInt = partialSums.last - partialSums(i)
def isRandLSumEqual(i: Int): Boolean = lSum(i) == rSum(i)
(0 until partialSums.length).find(isRandLSumEqual).getOrElse(-1)
}
$ include "seed7_05.s7i";
const array integer: numList is [] (-7, 1, 5, 2, -4, 3, 0);
const func array integer: equilibriumIndex (in array integer: elements) is func
result
var array integer: indexList is 0 times 0;
local
var integer: element is 0;
var integer: index is 0;
var integer: sum is 0;
var integer: subSum is 0;
var integer: count is 0;
begin
indexList := length(elements) times 0;
for element range elements do
sum +:= element;
end for;
for element key index range elements do
if 2 * subSum + element = sum then
incr(count);
indexList[count] := index;
end if;
subSum +:= element;
end for;
indexList := indexList[.. count];
end func;
const proc: main is func
local
var array integer: indexList is 0 times 0;
var integer: element is 0;
begin
indexList := equilibriumIndex(numList);
write("Found:");
for element range indexList do
write(" " <& element);
end for;
writeln;
end func;- Output:
Found: 4 7
func eq_index(nums) {
var (i, sum, sums) = (0, 0, Hash.new);
nums.each { |n|
sums{2*sum + n} := [] -> append(i++);
sum += n;
}
sums{sum} \\ [];
}
Test:
var indices = [
[-7, 1, 5, 2,-4, 3, 0],
[2, 4, 6],
[2, 9, 2],
[1,-1, 1,-1, 1,-1, 1],
]
for x in indices {
say ("%s => %s" % @|[x, eq_index(x)].map{.dump});
}
- Output:
[-7, 1, 5, 2, -4, 3, 0] => [3, 6] [2, 4, 6] => [] [2, 9, 2] => [1] [1, -1, 1, -1, 1, -1, 1] => [0, 1, 2, 3, 4, 5, 6]
extension Collection where Element: Numeric {
func equilibriumIndexes() -> [Index] {
guard !isEmpty else {
return []
}
let sumAll = reduce(0, +)
var ret = [Index]()
var sumLeft: Element = 0
var sumRight: Element
for i in indices {
sumRight = sumAll - sumLeft - self[i]
if sumLeft == sumRight {
ret.append(i)
}
sumLeft += self[i]
}
return ret
}
}
let arr = [-7, 1, 5, 2, -4, 3, 0]
print("Equilibrium indexes of \(arr): \(arr.equilibriumIndexes())")- Output:
Equilibrium indexes of [-7, 1, 5, 2, -4, 3, 0]: [3, 6]
proc listEquilibria {list} {
set after 0
foreach item $list {incr after $item}
set result {}
set idx 0
set before 0
foreach item $list {
incr after [expr {-$item}]
if {$after == $before} {
lappend result $idx
}
incr before $item
incr idx
}
return $result
}- Example of use
set testData {-7 1 5 2 -4 3 0}
puts Equilibria=[join [listEquilibria $testData] ", "]- Output:
Equilibria=3, 6
#import std
#import int
edex = num@yK33ySPtK33xtS2px; ~&nS+ *~ ==+ ~~r sum:-0
#cast %nL
example = edex <-7,1,5,2,-4,3,0>- Output:
<3,6>
Solution adopted from http://www.geeksforgeeks.org/equilibrium-index-of-an-array/ .
arr = Array(-7,1,5,2,-4,3,0)
WScript.StdOut.Write equilibrium(arr,UBound(arr))
WScript.StdOut.WriteLine
Function equilibrium(arr,n)
sum = 0
leftsum = 0
'find the sum of the whole array
For i = 0 To UBound(arr)
sum = sum + arr(i)
Next
For i = 0 To UBound(arr)
sum = sum - arr(i)
If leftsum = sum Then
equilibrium = equilibrium & i & ", "
End If
leftsum = leftsum + arr(i)
Next
End Function- Output:
Indices: 3, 6,
import "./fmt" for Fmt
var equilibrium = Fn.new { |a|
var len = a.count
var equi = []
if (len == 0) return equi // sequence has no indices at all
var rsum = a.reduce { |acc, x| acc + x }
var lsum = 0
for (i in 0...len) {
rsum = rsum - a[i]
if (rsum == lsum) equi.add(i)
lsum = lsum + a[i]
}
return equi
}
var tests = [
[-7, 1, 5, 2, -4, 3, 0],
[2, 4, 6],
[2, 9, 2],
[1, -1, 1, -1, 1, -1, 1],
[1],
[]
]
System.print("The equilibrium indices for the following sequences are:\n")
for (test in tests) {
Fmt.print("$24n -> $n", test, equilibrium.call(test))
}- Output:
The equilibrium indices for the following sequences are:
[-7, 1, 5, 2, -4, 3, 0] -> [3, 6]
[2, 4, 6] -> []
[2, 9, 2] -> [1]
[1, -1, 1, -1, 1, -1, 1] -> [0, 1, 2, 3, 4, 5, 6]
[1] -> [0]
[] -> []
code Ran=1, ChOut=8, IntOut=11;
def Size = 1_000_000;
int I, S, A(Size), Hi(Size), Lo(Size);
[for I:= 0 to Size-1 do A(I):= Ran(100) - 50;
S:= 0;
for I:= 0 to Size-1 do [S:= S+A(I); Lo(I):= S];
S:= 0;
for I:= Size-1 downto 0 do [S:= S+A(I); Hi(I):= S];
for I:= 0 to Size-1 do
if Lo(I) = Hi(I) then [IntOut(0, I); ChOut(0, ^ )];
]- Output:
502910 504929 508168
Yorick arrays are 1-based so the output of this program will be shifted up by one compared to solutions in languages with 0-based arrays.
func equilibrium_indices(A) {
return where(A(psum) == A(::-1)(psum)(::-1));
}- Example interactive usage:
> equilibrium_indices([-7, 1, 5, 2, -4, 3, 0]) [4,7]
fcn equilibrium(lst){ // two pass
reg acc=List(), left=0,right=lst.sum(0),i=0;
foreach x in (lst){
right-=x;
if(left==right) acc.write(i);
i+=1; left+=x;
}
acc
}fcn equilibrium(lst){ // lst should immutable, n^2
(0).filter(lst.len(),'wrap(n){ lst[0,n].sum(0) == lst[n+1,*].sum(0) })
}If the input list is immutable, no new lists are generated (other than accumulating the result).
equilibrium(T(-7, 1, 5, 2, -4, 3, 0)).println();- Output:
L(3,6)
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