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In this post, you will find the solution for the **Merge Intervals** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Merge Intervals– LeetCode Problem

Merge Intervals– LeetCode Problem

**Problem:**

Given an array of `intervals`

where `intervals[i] = [start`

, merge all overlapping intervals, and return _{i}, end_{i}]*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] overlaps, 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 <= 10`

^{4}`intervals[i].length == 2`

`0 <= start`

_{i}<= end_{i}<= 10^{4}

Merge Intervals– LeetCode Solutions

Merge Intervals in C++:

class Solution { public: vector> merge(vector >& intervals) { vector > ans; sort(begin(intervals), end(intervals)); for (const auto& interval : intervals) if (ans.empty() || ans.back()[1] < interval[0]) ans.push_back(interval); else ans.back()[1] = max(ans.back()[1], interval[1]); return ans; } };

Merge Intervals in Java:

class Solution { public int[][] merge(int[][] intervals) { Listans = new ArrayList<>(); Arrays.sort(intervals, (a, b) -> (a[0] - b[0])); for (int[] interval : intervals) if (ans.isEmpty() || ans.get(ans.size() - 1)[1] < interval[0]) ans.add(interval); else ans.get(ans.size() - 1)[1] = Math.max(ans.get(ans.size() - 1)[1], interval[1]); return ans.toArray(new int[ans.size()][]); } }

Merge Intervals in Python:

class Solution: def merge(self, intervals: List[List[int]]) -> List[List[int]]: ans = [] for interval in sorted(intervals): if not ans or ans[-1][1] < interval[0]: ans.append(interval) else: ans[-1][1] = max(ans[-1][1], interval[1]) return ans