# 452. Minimum Number of Arrows to Burst Balloons

### Description

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*.

### Constraints

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

### Approach

### Links

* GeeksforGeeks
* [Leetcode](https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/)
* ProgramCreek
* YouTube

### **Examples**

{% tabs %}
{% tab title="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).
{% endtab %}

{% tab title="Example 2" %}
**Input:** points = \[\[1, 2], \[3, 4], \[5, 6], \[7, 8]]

**Output:** 4
{% endtab %}

{% tab title="Example 3" %}
**Input:** points = \[\[1, 2], \[2, 3], \[3, 4], \[4, 5]]

**Output:** 2
{% endtab %}

{% tab title="Example 4" %}
**Input:** points = \[\[1, 2]]

**Output:** 1
{% endtab %}

{% tab title="Example 5" %}
**Input:** points = \[\[2, 3], \[2, 3]]

**Output:** 1
{% endtab %}
{% endtabs %}

### **Solutions**

{% tabs %}
{% tab title="Solution 1" %}

```java
/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int findMinArrowShots(int[][] points) {
        if(points == null || points.length == 0) return 0;
        
        Arrays.sort(points, (a, b) -> (a[1] > b[1])? 1: -1);
        
        int arrowCount = 1,
            arrowPos = points[0][1];
        
        for(int i = 1; i < points.length; i++) {
            if(points[i][0] > arrowPos) {
                arrowCount++;
                arrowPos = points[i][1];
            }
        }
        
        return arrowCount;
    }
}
```

{% endtab %}
{% endtabs %}

### **Follow up**

*