insert-interval.py

from __future__ import print_function
# Time:  O(n)
# Space: O(1)

# 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].

# Definition for an interval.
class Interval:
    def __init__(self, s=0, e=0):
        self.start = s
        self.end = e

    def __repr__(self):
        return "[{}, {}]".format(self.start, self.end)


class Solution(object):
    def insert(self, intervals, newInterval):
        """
        :type intervals: List[Interval]
        :type newInterval: Interval
        :rtype: List[Interval]
        """
        result = []
        i = 0
        while i < len(intervals) and newInterval.start > intervals[i].end:
            result += intervals[i],
            i += 1
        while i < len(intervals) and newInterval.end >= intervals[i].start:
            newInterval = Interval(min(newInterval.start, intervals[i].start), \
                                   max(newInterval.end, intervals[i].end))
            i += 1
        result += newInterval,
        result += intervals[i:]
        return result


if __name__ == "__main__":
    print(Solution().insert([Interval(1, 2), Interval(3, 5), Interval(6, 7), Interval(8, 10), Interval(12, 16)], Interval(4, 9)))