notes/OJ notes/pages/Leetcode Non-Overlapping-Intervals.md

# Leetcode Non-Overlapping-Intervals

2022-09-03 15:34

> ##### Algorithms:
>
> #algorithm #greedy
>
> ##### Data structures:
>
> #DS #array
>
> ##### Difficulty:
>
> #coding_problem #difficulty_medium
>
> ##### Additional tags:
>
> #<platforms-like-leetcode> #<list-to-be-added>
>
> ##### Revisions:
>
> N/A

##### Links:

- [Link to problem](https://leetcode.com/problems/non-overlapping-intervals/)

---

### Problem

Given an array of intervals `intervals` where `intervals[i] = [starti, endi]`, return _the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping_.

#### Examples

**Example 1:**

**Input:** intervals = [[1,2],[2,3],[3,4],[1,3]]
**Output:** 1
**Explanation:** [1,3] can be removed and the rest of the intervals are non-overlapping.

**Example 2:**

**Input:** intervals = [[1,2],[1,2],[1,2]]
**Output:** 2
**Explanation:** You need to remove two [1,2] to make the rest of the intervals non-overlapping.

**Example 3:**

**Input:** intervals = [[1,2],[2,3]]
**Output:** 0
**Explanation:** You don't need to remove any of the intervals since they're already non-overlapping.

#### Constraints

- `1 <= intervals.length <= 105`
- `intervals[i].length == 2`
- `-5 * 104 <= starti < endi <= 5 * 104`

### Thoughts

> [!summary]
> This is a #greedy problem, similar to [[Leetcode Merge-Intervals]]

#### Key concept:

first sort the intervals.

pick the intervals with smallest end, which will allow us to
make more room for following ones.

#tip: Use `&` to reference vectors in comp.function to save
time.

### Solution

```cpp
class Solution {
  static bool comp(vector<int> &a, vector<int> &b) { return a[1] < b[1]; }

public:
  int eraseOverlapIntervals(vector<vector<int>> &intervals) {

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

    // Using this var to skip overlapping intervals.
    auto before = intervals[0];

    int ans = 0;

    for (int i = 1, size = intervals.size(); i < size; i++) {
      if (intervals[i][0] < before[1]) {
        ans++;
      } else {
        before = intervals[i];
      }
    }

    return ans;
  }
};
```