# 452. Minimum Number of Arrows to Burst Balloons

#### Medium

***

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*.

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**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].
```

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**Constraints:**

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

```python
class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        points.sort(key = lambda x: x[1])
        # print(points)
        common = []
        for point in points:
            if len(common) == 0:
                common.append(point)
                continue
            last = common[-1]
            if last[1] >= point[0]:
                temp = [max(point[0], last[0]), min(point[1], last[1])]
                if temp[0] <= last[0] or temp[1] <= last[1]:
                    common[-1] = temp
                else:
                    common.append(temp)
            else:
                common.append(point)
            # print('->', common)
        return len(common)
```