# Merge Intervals Leetcode Solution

In this post, we are going to solve the Merge Intervals Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

## Problem

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`

Now, lets see the leetcode solution of Merge Intervals Leetcode Solution.

### Merge Intervals Leetcode Solution 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
```

### Merge Intervals Leetcode Solutionin CPP

```class Solution {
public:
vector<vector<int>> merge(vector<vector<int>>& intervals) {
vector<vector<int>> ans;

sort(begin(intervals), end(intervals));

for (const vector<int>& 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 Leetcode Solution in Java

```class Solution {
public int[][] merge(int[][] intervals) {
List<int[]> ans = 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])