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# Minimum Number of Arrows to Burst Balloons

## [Minimum Number of Arrows to Burst Balloons](https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons)

There are some spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter, and hence the x-coordinates of start and end of the diameter suffice. The start is always smaller than the end.

An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with `xstart` and `xend` bursts by an arrow shot at `x` if `xstart ≤ x ≤ xend`. There is no limit to the number of arrows that can be shot. An arrow once shot keeps traveling up infinitely.

Given an array `points` where `points[i] = [xstart, xend]`, return *the minimum number of arrows that must be shot to burst all balloons*.

**Example 1:**

```

Input: points = [[10,16],[2,8],[1,6],[7,12]]
Output: 2
Explanation: One way is to shoot one arrow for example at x = 6 (bursting the balloons [2,8] and [1,6]) and another arrow at x = 11 (bursting the other two balloons).
```

**Example 2:**

```

Input: points = [[1,2],[3,4],[5,6],[7,8]]
Output: 4
```

**Example 3:**

```

Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2
```

**Example 4:**

```

Input: points = [[1,2]]
Output: 1
```

**Example 5:**

```

Input: points = [[2,3],[2,3]]
Output: 1
```

**Constraints:**

* `0 <= points.length <= 104`
* `points[i].length == 2`
* `-231 <= xstart < xend <= 231 - 1`

## Solutions

### 🧠 Cpp

```cpp
class Solution
{
    enum {START, END};

public:
    //O(n log n) solution, because of sorting
    //sort intervals by END, (wider first if ENDs are the same)
    //and check for overlaps with current->at(END) < next_iter->at(START)
    int findMinArrowShots(vector<vector<int>>& points)
    {
        //sort by END value, if equal: put wider intervals first
        std::sort(begin(points), end(points),
                  [](auto a, auto b)
                  {
                      return a[END] == b[END] ? 
                      a[START] > b[START] :
                      a[END] < b[END];
                  });

        int arrow_num = 0;
        for(auto current = begin(points); current != end(points);)
        {
            //count all intervals that overlaps with current
            //still O(n) because we moving forward in vector
            for(auto next_iter = next(current); ; ++next_iter)
            {
                if(next_iter == end(points) || 
                   current->at(END) < next_iter->at(START))
                {
                    arrow_num++;
                    current = next_iter;
                    break;
                }
                //else there is overlap with current
            }
        }

        return arrow_num;
    }
};
```