# Leetcode Merge-Intervals

id:: 642edfd5-4367-495d-b836-510b0deaa3bf

#### 2022-09-01 14:04

> ##### Algorithms:
>
> #algorithm #sort
>
> ##### Data structures:
>
> #DS #array
>
> ##### Difficulty:
>
> #coding_problems #difficulty_medium
>
> ##### Additional tags:
>
> #leetcode
>
> ##### Revisions:
>
> N/A

##### Links:

- [Link to problem](https://leetcode.com/problems/merge-intervals/)
- [Solution ans Explanation](https://leetcode.com/problems/merge-intervals/solution/)

  ***

### 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_.

- #### Examples
- ```
  **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`

### Thoughts

> [!summary]
> This is a generic array problem.

#### Situations to consider:

- The intervals can be unordered
- The first interval
- `[0, 3] [0, 1]` are adjacent and overlapped.

  To solve the situations, sort first, and use `max` function to solve the 3rd solution.

### Solution

```cpp
class Solution {
public:
vector<vector<int>> merge(vector<vector<int>> &intervals) {
// sort first, so that data are continious
vector<vector<int>> ans;

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

for (auto interval : intervals) {
  if (ans.empty() || ans.back()[1] < interval[0]) {
    ans.push_back(interval);
  } else {
    ans.back()[1] = max(interval[1], ans.back()[1]);
  }
}

return ans;
}
};
```
