Skip to content
Programmingoneonone
Programmingoneonone
  • CS Subjects
    • Internet of Things (IoT)
    • Digital Communication
    • Human Values
    • Cybersecurity
  • Programming Tutorials
    • C Programming
    • Data structures and Algorithms
    • 100+ Java Programs
    • 100+ C Programs
  • HackerRank Solutions
    • HackerRank Algorithms Solutions
    • HackerRank C problems solutions
    • HackerRank C++ problems solutions
    • HackerRank Java problems solutions
    • HackerRank Python problems solutions
Programmingoneonone
Programmingoneonone

Leetcode Merge Intervals problem solution

YASH PAL, 31 July 202418 January 2026

In this Leetcode Merge Intervals problem solution, we have 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.

Leetcode Merge Intervals problem solution

Leetcode Merge Intervals problem solution in Python.

def merge(self, intervals):
    if not intervals:
        return []
    
    intervals.sort(key = lambda x: x[0])
    n = len(intervals)
    preS = intervals[0][0]
    preE = intervals[0][1]
    res = []
    
    for i in range(1, n):
        curS = intervals[i][0]
        curE = intervals[i][1]
        if curS <= preE:
            preE = max(curE, preE)
        else:
            res.append([preS, preE])
            preS = curS
            preE = curE
            
    res.append([preS, preE])

    return res

Merge Intervals problem solution in Java.

public int[][] merge(int[][] intervals) {
   if(intervals.length == 1 || intervals.length == 0) return intervals;
    Arrays.sort(intervals,new Comparator<int[]>(){
        public int compare(int[] o1, int[] o2){
            if(o1[0] == o2[0])
                return o1[1]-o2[1];
            else return o1[0]-o2[0];
        }
    });
    int start = 0;
    int end = 1;

    while(end < intervals.length) {
        if(intervals[start][1] >= intervals[end][0]){
            if(intervals[end][1] > intervals[start][1])
                intervals[start][1] = intervals[end][1];
            
        } else {
            start++;
            if(start != end){
                intervals[start][0] = intervals[end][0];
                intervals[start][1] = intervals[end][1];
            }
            
        }
        end++;
    }
    int[][] res = new int[start+1][2];
    for(int i=0;i<=start;i++)
        res[i] = intervals[i];
  


    return res;
}

Problem solution in C++.

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        vector<vector<int > > v;
        sort(intervals.begin(), intervals.end());
        
        for(auto i: intervals) {
            int st = i[0], en = i[1];
            
            if(v.size() && st <= v.back()[1]) {
                st = v.back()[0];
                en = max(i[1], v.back()[1]);

                v.pop_back();
            }
            
            v.push_back({st, en});
        }
        
        return v;
    }
};

Problem solution in C.

int cmp(void* a,void* b){
        if(((int**)a)[0][0]==((int**)b)[0][0]){
            return ((int**)a)[0][1]-((int**)b)[0][1];
        }
        return ((int**)a)[0][0]-((int**)b)[0][0];
    }

    int** merge(int** intervals, int intervalsSize, int* intervalsColSize, int* returnSize, int** returnColumnSizes){
        *returnSize=0;
        if(intervalsSize==0){
            return NULL;
        }
        if(intervalsSize==1){
            *returnSize=1;
            return intervals;
        }
        qsort(intervals,intervalsSize,sizeof(intervals[0]),cmp);
        int** ret=(int**)malloc(intervalsSize*sizeof(int*));
        int* tmp=NULL;
        int i=0;
        tmp=intervals[0];
        for(i=1;i<intervalsSize;i++){
            if(tmp[1]>=intervals[i][0]){
                tmp[1]=tmp[1]>intervals[i][1] ? tmp[1]: intervals[i][1];
            }
            else{    
                ret[(*returnSize)++]=tmp;
                tmp=intervals[i];  
            }
        }
        if(*returnSize>0 && ret[(*returnSize-1)][1]<tmp[0]){
            ret[(*returnSize)++]=tmp;
        }
        if(*returnSize==0){
            ret[(*returnSize)++]=tmp;
        }
        returnColumnSizes[0]=(int*)malloc((*returnSize)*sizeof(int));
        for(int i=0;i<(*returnSize);i++){
            returnColumnSizes[0][i]=2;
        }
        return ret;
    }

coding problems solutions Leetcode Problems Solutions Leetcode

Post navigation

Previous post
Next post

Leave a Reply

Your email address will not be published. Required fields are marked *

CLOSE ADS
CLOSE ADS

Pages

  • About US
  • Contact US
  • Privacy Policy

Follow US

  • YouTube
  • LinkedIn
  • Facebook
  • Pinterest
  • Instagram
©2026 Programmingoneonone | WordPress Theme by SuperbThemes