FilesExpand file tree

 master

/

MinimumWindowSubstring.java

Copy path

BlameMore file actions

BlameMore file actions

Latest commit

 

History

History

History

53 lines (50 loc) · 1.3 KB

 master

/

MinimumWindowSubstring.java

Top

File metadata and controls

• Code
• Blame

53 lines (50 loc) · 1.3 KB

Raw

Copy raw file

Download raw file

Open symbols panel

Edit and raw actions

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

/**

* Given a string S and a string T, find the minimum window in S which will

* contain all the characters in T in complexity O(n).

*

* For example, S = "ADOBECODEBANC" T = "ABC" Minimum window is "BANC".

*

* Note:

*

* If there is no such window in S that covers all characters in T, return the emtpy string "".

*

* If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

*/

public class MinimumWindowSubstring {

public String minWindow(String S, String T) {

int[] hasFound = new int[256];

int[] needFound = new int[256];

int diffCount = T.length();

int length = S.length();

String window = "";

int size = Integer.MAX_VALUE;

if (length == 0 || diffCount == 0)

return window;

for (int l = 0; l < diffCount; l++) {

needFound[T.charAt(l)]++;

}

int i = 0, j = 0;

while (j < length) {

char c = S.charAt(j);

if (needFound[c] > 0 && ++hasFound[c] <= needFound[c]) {

diffCount--;

}

if (diffCount == 0) {

while (i <= j) {

char h = S.charAt(i);

i++;

if (needFound[h] > 0 && --hasFound[h] < needFound[h]) {

if (j - i + 1 < size) {

size = j - i + 1;

window = S.substring(i - 1, j + 1);

}

diffCount++;

break;

}

}

}

j++;

}

return window;

}

}