# Leetcode 452. Minimum Number of Arrows to Burst Balloons

`Input: points = [[10,16],[2,8],[1,6],[7,12]]Output: 2Explanation: 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].`
`Input: points = [[1,2],[3,4],[5,6],[7,8]]Output: 4Explanation: One arrow needs to be shot for each balloon for a total of 4 arrows.`
`Input: points = [[1,2],[2,3],[3,4],[4,5]]Output: 2Explanation: 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].`
• `1 <= points.length <= 105`
• `points[i].length == 2`
• `-231 <= xstart < xend <= 231 - 1`
1. it’s like sliding window problem, and we can use greedy algorithm to solve
2. sort the array by the ending point, and loop the array
3. if next ballon starting point before previous ending point, we can shoot them together.
4. to avoid overflow, use Integer.compare(a, b), instead of (a-b)

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