# Posts published in “Interval Merge”

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] 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 <= 104`
• `intervals[i].length == 2`
• `0 <= starti <= endi <= 104`

Idea:

• First step (Important): sort the intervals by the first number
• Use result array to store the first interval
• Compare the last interval of the result array vs. new coming interval
• new interval > last interval of the result array : store the new interval into result array
• new interval < last interval of the result array : store the interval that has longer range

Solution:

``````/**
* @param {number[][]} intervals
* @return {number[][]}
*/
var merge = function(intervals) {
intervals.sort(([a, b], [c, d]) => a - c);
let res = [];
for (let interval of intervals) {
if (res.length === 0 || interval > res[res.length - 1]) {
res.push(interval.concat());
} else {
res[res.length - 1] = Math.max(interval, res[res.length - 1]);
}
}
return res;
};``````