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#!/usr/bin/env python
'''
Leetcode: Insert Interval.
Given a set of [[[non-overlapping]]] intervals, insert a new interval into the intervals (merge if necessary). You may assume that the intervals were initially sorted according to their start times.
Example 1: Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9].
Example 2: Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16]. This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10].
'''
from __future__ import division
import random
# interval tree? segment tree?
# [1,3],[6,9] --> index
# [-inf. 1)->0; [1,3]->1; (3,6)->2; [6,9]->3; (9,inf.)->4
def insert_interval(intvs, new_intv):
left, right = new_intv
ret_intvs = []
new_intv_inserted = False
for a,b in intvs:
if left > b: # not overlap at all
ret_intvs.append([a,b])
elif right < a:
if not new_intv_inserted: # not overlap at all
ret_intvs.append([left,right])
new_intv_inserted = True
ret_intvs.append([a,b])
else:
# starting/ending point in the interval
if a <= left <= b: left = a
if a <= right <= b: right = b
# leftover
if not new_intv_inserted:
ret_intvs.append([left,right])
return ret_intvs
## in-place
def insert_interval2(intvs, new_intv):
intvs = sorted(intvs, key=lambda x:x[0])
i = 0
while i < len(intvs):
cur = intvs[i]
if new_intv[1] < cur[0] or new_intv[0] > cur[1]:
i += 1
else:
left = min(cur[0], new_intv[0])
right = max(cur[1], new_intv[1])
new_intv = [left, right]
del intvs[i]
intvs.append(new_intv)
return intvs
if __name__ == '__main__':
intvs = [[1,2],[3,5],[6,7],[8,10],[12,16]]; new_intv = [4,9]
print insert_interval2(intvs, new_intv)
intvs = [[1,10],[12,16]]; new_intv = [2,8]
print insert_interval2(intvs, new_intv)
intvs = [[1,10],[12,16]]; new_intv = [-5,0]
print insert_interval2(intvs, new_intv)
intvs = [[1,10],[12,16]]; new_intv = [20,20]
print insert_interval2(intvs, new_intv)
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