# 452. Minimum Number of Arrows to Burst Balloons | LeetCode

## 452. Minimum Number of Arrows to Burst Balloons

There are some spherical balloons taped onto a flat wall that represents the XY-plane. The balloons are represented as a 2D integer array `points` where `points[i] = [xstart, xend]` denotes a balloon whose horizontal diameter stretches between `xstart` and `xend`. You do not know the exact y-coordinates of the balloons.

Arrows can be shot up directly vertically (in the positive y-direction) from different points along the x-axis. A balloon with `xstart` and `xend` is burst by an arrow shot at `x` if `xstart <= x <= xend`. There is no limit to the number of arrows that can be shot. A shot arrow keeps traveling up infinitely, bursting any balloons in its path.

Given the array `points`, 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: The balloons can be burst by 2 arrows:
- Shoot an arrow at x = 6, bursting the balloons [2,8] and [1,6].
- Shoot an arrow at x = 11, bursting the balloons [10,16] and [7,12].
```

Example 2:

```Input: points = [[1,2],[3,4],[5,6],[7,8]]
Output: 4
Explanation: One arrow needs to be shot for each balloon for a total of 4 arrows.
```

Example 3:

```Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2
Explanation: The balloons can be burst by 2 arrows:
- Shoot an arrow at x = 2, bursting the balloons [1,2] and [2,3].
- Shoot an arrow at x = 4, bursting the balloons [3,4] and [4,5].
```

Constraints:

• `1 <= points.length <= 105`
• `points[i].length == 2`
• `-231 <= xstart < xend <= 231 - 1`

### The solution to the above question

bool cmp(vector<int>& a, vector<int>& b) {return a[1] < b[1];}
class Solution {
public:
int findMinArrowShots(vector<vector<int>>& v) {
sort(v.begin(),v.end(),cmp);
vector<int>temp=v[0];
vector<vector<int>>ans;
for(int i=1; i<v.size(); i++){
if(temp[1]>=v[i][0] && temp[1]<=v[i][1]){
temp[0]=min(v[i][0],temp[1]);
temp[1]=max(v[i][0],temp[1]);
}
else if(v[i][0]>temp[1]){
ans.push_back(temp);
temp.clear();
temp=v[i];
}
}

ans.push_back(temp);
for(int i=0; i<ans.size(); i++){
cout<<ans[i][0]<<" "<<ans[i][1]<<endl;
}
return ans.size();
}
};