# Day 37 — Merge Intervals

Given a collection of intervals, merge all overlapping intervals.

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## Introduction🛹

Hey Guys, Today is day 37 *of the challenge that I took. Wherein I will be solving every day for 100 days the *programming questions that have been asked in** previous interviews.**

You have a bonus at the end if you keep reading. You can find out the companies that have asked these questions in real interviews.

All these problems are taken from the following e-book. 🎓

This is completely free 🆓 if you have an amazon kindle subscription.

This** e-book contains 100 coding problems** that have been asked in top tech interview questions. It also has a guide to solving *all the problems in **200+ ways. *These problems I assure you** has been asked in previous interviews.

You have to decide whether you want to go** unprepared *for a tech interview or go ahead and **quick search for this guide *to solve the 100 problems.

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Note :_ this e-book only contains the _links __to the__solutions.

## AIM🏹

Given a collection of intervals, merge *all *overlapping intervals.

## Example🕶

``````Input: [[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].``````

## Code👇

``````class Solution {
public int[][] merge(int[][] intervals) {
if (intervals.length == 0) {
return new int[0][0];
}

Arrays.sort(intervals, (i1, i2) -> i1[0] - i2[0]);
List<int[]> list = new ArrayList<>();
for (int i = 1; i < intervals.length; i++) {
int[] lastItem = list.get(list.size() - 1);

if (intervals[i][0] > lastItem[1]) {
} else if (intervals[i][1] > lastItem[1]) {
list.remove(list.size()-1);
}
}

int ans[][]= new int[list.size()][2];

for(int i=0; i< list.size(); i++)
{

int arr[] = list.get(i);
ans[i][0] = arr[0];
ans[i][1] = arr[1];

}
// int ans[] = new in[2][2];
return ans;
}
}``````

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