57. Insert Interval

Problem Statement

You are given an array of non-overlapping intervals intervals where intervals[i] = [starti, endi] represent the start and the end of the ith interval and intervals is sorted in ascending order by starti. You are also given an interval newInterval = [start, end] that represents the start and end of another interval.

Insert newInterval into intervals such that intervals is still sorted in ascending order by starti and intervals still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return intervals after the insertion.

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].

Constraints:

0 <= intervals.length <= 104
intervals[i].length == 2
0 <= starti <= endi <= 105
intervals is sorted by starti in ascending order.
newInterval.length == 2
0 <= start <= end <= 105

Intuition

Approach:

First we account for non-overlap at start
Then overlap if any 
Then overlap case

Links

https://leetcode.com/problems/insert-interval/description/

Video Links

Approach 1:

C++

class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        int n = intervals.size(), i = 0;
        vector<vector<int>> ans;
        // Case 1 : No Overlapping case before the merge intervals
        while(i < n && intervals[i][1] < newInterval[0]){
            ans.push_back(intervals[i]);
            i++;
        }
        // Case 2 : Overlapping Case and merging of intervals
        while(i < n && newInterval[1] >= intervals[i][0]){
            newInterval[0] = min(newInterval[0], intervals[i][0]);
            newInterval[1] = max(newInterval[1], intervals[i][1]);
            i++;
        }
        ans.push_back(newInterval);
        // Case 3 : No overlapping of interval after merge
        while(i < n){
            ans.push_back(intervals[i]);
            i++;
        }
        return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++