LeetCode 57. Insert Interval

Description

https://leetcode.com/problems/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:

Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
Output: [[1,5],[6,9]]

Example 2:

Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
Output: [[1,2],[3,10],[12,16]]
Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].

Example 3:

Input: intervals = [], newInterval = [5,7]
Output: [[5,7]]

Example 4:

Input: intervals = [[1,5]], newInterval = [2,3]
Output: [[1,5]]

Example 5:

Input: intervals = [[1,5]], newInterval = [2,7]
Output: [[1,7]]

Constraints:

  • 0 <= intervals.length <= 104
  • intervals[i].length == 2
  • 0 <= intervals[i][0] <= intervals[i][1] <= 105
  • intervals is sorted by intervals[i][0] in ascending order.
  • newInterval.length == 2
  • 0 <= newInterval[0] <= newInterval[1] <= 105

Explanation

We can insert the new interval first then merge all intervals.

Python Solution

class Solution:
    def insert(self, intervals: List[List[int]], newInterval: List[int]) -> List[List[int]]:
        
        intervals.append(newInterval)
        
        intervals = sorted(intervals, key=lambda interval:interval[0])
        
        results = []
        
        prev = intervals[0]
        
        for interval in intervals[1:]:   
            if interval[0] <= prev[1]:
                prev[1] = max(interval[1], prev[1])
            else:
                results.append(prev)
                prev = interval
                
        results.append(prev)
            
        return results
  • Time Complexity: O(Nlog(N)).
  • Space Complexity: O(N).

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