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# Merge Overlapping Intervals

{% hint style="info" %}

### Merge Overlapping Intervals

Write a function that takes in a non-empty array of arbitrary intervals, merges any overlapping intervals, and returns the new intervals in no particular order.

Each interval interval is an array of two integers, with interval\[0] as the start of the interval and interval\[1] as the end of the interval.

Note that back-to-back intervals aren't considered to be overlapping. For example, \[1, 5] and \[6, 7] aren't overlapping; however, \[1, 6] and \[6, 7] *are* indeed overlapping.

Also note that the start of any particular interval will always be less than or equal to the end of that interval.

#### Sample Input

```
intervals = [[1, 2], [3, 5], [4, 7], [6, 8], [9, 10]]
```

#### Sample Output

```
[[1, 2], [3, 8], [9, 10]]
// Merge the intervals [3, 5], [4, 7], and [6, 8].
// The intervals could be ordered differently.
```

{% endhint %}

{% tabs %}
{% tab title="Solution 1" %}
{% code lineNumbers="true" %}

```javascript
// O(nlog(n)) time | O(n) space
function mergeOverlappingIntervals(intervals) {
  const sortedIntervals = intervals.sort((a,b) => a[0] - b[0])
  const mergedIntervals = [];
  let currentInterval = sortedIntervals[0];
  mergedIntervals.push(currentInterval);

  for(const nextInterval of sortedIntervals) {
    const [_, currentIntervalEnd] = currentInterval;
    const [nextIntervalStart, nextIntervalEnd] = nextInterval;
    if (currentIntervalEnd >= nextIntervalStart) {
      currentInterval[1] = Math.max(currentIntervalEnd, nextIntervalEnd);
    } else {
      currentInterval = nextInterval;
      mergedIntervals.push(currentInterval)
    }
  }
  return mergedIntervals;
}

// Do not edit the line below.
exports.mergeOverlappingIntervals = mergeOverlappingIntervals;

```

{% endcode %}
{% endtab %}
{% endtabs %}