56. Merge Intervals

Problem Statement

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

Intuition

Approach:

Merge the intervals as we go along, sort based on start value
Then modify the array

Edge case:
1,5  : 2,4
We take max of 5,4 when we modify end time

Links

https://leetcode.com/problems/merge-intervals/description/

Video Links

https://www.youtube.com/watch?v=44H3cEC2fFM&ab_channel=NeetCode

Approach 1:

C++

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        int n = intervals.size();
        vector<vector<int>> ans;
        sort(intervals.begin(), intervals.end());
        ans.push_back(intervals[0]);

        for(int i=1; i<n; i++){
            int size = ans.size();
            if( intervals[i][0] <= ans[size-1][1] ){
                ans[size-1][1] = max(ans[size-1][1], intervals[i][1]);
                //1,5 : 2,4 To account this we take max
            }
            else{
                ans.push_back(intervals[i]);
            }
        }

        return ans;
    }
};

Approach 2:

C++

Approach 3:

C++

Approach 4:

C++

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