Medium

Question

Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return an array of the non-overlapping intervals that cover all the intervals in the input.

Example 1:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlap, merge them into [1,6].

Example 2:

Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

Constraints:

• 1 <= intervals.length <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 104

Hint

Check the start and end of consecutive intervals in a sorted list

Example: [[i, j],[p, q]] compare i&p and j&q

Solution

My Solution:

import heapq
class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        stack = []
        intervals.sort(key = lambda x: x[0])
        for i,j in intervals:
            if stack and stack[-1][0] <= i <= stack[-1][1]:
                p,q = stack.pop()
                stack.append([min(i,p), max(j,q)])
            else:
                stack.append([i,j])
        return stack

Time Complexity: O(nlogn) # sorting takes O(nlogn) and the O(n) for iterating through intervals

Space Complexity: O(n) # since we have a stack of size n in worst case