Here, We see Merge Intervals LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Merge Intervals LeetCode Solution

Problem Statement

Given an array of intervals where intervals[i] = [start_i, end_i], 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.

Merge Intervals Solution C++

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        if(intervals.size()<=1) return intervals;
        sort(intervals.begin(), intervals.end());
        vector<vector<int>> output;
        output.push_back(intervals[0]);
        for(int i=1; i<intervals.size(); i++) {
            if(output.back()[1] >= intervals[i][0]) output.back()[1] = max(output.back()[1] , intervals[i][1]);
            else output.push_back(intervals[i]); 
        }
        return output;        
    }
};Code language: C++ (cpp)

Merge Intervals Solution Java

class Solution {
    public int[][] merge(int[][] intervals) {
        if(intervals == null || intervals.length == 0)
            return intervals;
        Arrays.sort(intervals, (a, b) -> Integer.compare(a[0], b[0]));
        LinkedList<int[]> mergedIntervals = new LinkedList<>();
        for(int[] curr : intervals) {
            if(mergedIntervals.isEmpty() || mergedIntervals.getLast()[1] < curr[0])
                mergedIntervals.add(curr);
            else 
                mergedIntervals.getLast()[1] = Math.max(mergedIntervals.getLast()[1], curr[1]);
        }
        return mergedIntervals.toArray(new int[0][]);        
    }
}
Code language: Java (java)

Merge Intervals Solution JavaScript

var merge = function(intervals) {
    if(!intervals.length) return [];
    intervals.sort((a, b) => a[0] - b[0]);
    
    const result = [intervals[0]];
    
    for(let [start, end] of intervals) {
        const endPrev = result.at(-1)[1]
        if(start <= endPrev) result.at(-1)[1] = Math.max(end, endPrev);
        else result.push([start, end]);
    }
    return result; 
};
Code language: JavaScript (javascript)

Merge Intervals Solution Python

class Solution(object):
    def merge(self, intervals):
        START, END = 0, 1
        result = []
        intervals.sort( key = lambda x: (x[START], x[END] ) ) 
        for interval in intervals:
            if not result or ( result[-1][END] < interval[START] ):
                result.append( interval )
            else:
                result[-1][END] = max(result[-1][END], interval[END])
        return resultCode language: Python (python)