# 57. Insert Interval

## Problem Statement

<br>

You are given an array of non-overlapping intervals `intervals` where `intervals[i] = [starti, endi]` represent the start and the end of the `ith` interval and `intervals` is sorted in ascending order by `starti`. You are also given an interval `newInterval = [start, end]` that represents the start and end of another interval.

Insert `newInterval` into `intervals` such that `intervals` is still sorted in ascending order by `starti` and `intervals` still does not have any overlapping intervals (merge overlapping intervals if necessary).

Return `intervals` *after the insertion*.

&#x20;

**Example 1:**

<pre><code><strong>Input: intervals = [[1,3],[6,9]], newInterval = [2,5]
</strong><strong>Output: [[1,5],[6,9]]
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: intervals = [[1,2],[3,5],[6,7],[8,10],[12,16]], newInterval = [4,8]
</strong><strong>Output: [[1,2],[3,10],[12,16]]
</strong><strong>Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
</strong></code></pre>

&#x20;

**Constraints:**

* `0 <= intervals.length <= 104`
* `intervals[i].length == 2`
* `0 <= starti <= endi <= 105`
* `intervals` is sorted by `starti` in **ascending** order.
* `newInterval.length == 2`
* `0 <= start <= end <= 105`

## Intuition

```
Approach:

First we account for non-overlap at start
Then overlap if any 
Then overlap case
```

### Links

<https://leetcode.com/problems/insert-interval/description/>

### Video Links

### Approach 1:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    vector<vector<int>> insert(vector<vector<int>>& intervals, vector<int>& newInterval) {
        int n = intervals.size(), i = 0;
        vector<vector<int>> ans;
        // Case 1 : No Overlapping case before the merge intervals
        while(i < n && intervals[i][1] < newInterval[0]){
            ans.push_back(intervals[i]);
            i++;
        }
        // Case 2 : Overlapping Case and merging of intervals
        while(i < n && newInterval[1] >= intervals[i][0]){
            newInterval[0] = min(newInterval[0], intervals[i][0]);
            newInterval[1] = max(newInterval[1], intervals[i][1]);
            i++;
        }
        ans.push_back(newInterval);
        // Case 3 : No overlapping of interval after merge
        while(i < n){
            ans.push_back(intervals[i]);
            i++;
        }
        return ans;
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###


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