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 idea of this problem is finding overlapping between different points, then merge and count. To do so, we need to sort the points by the first element, then we don’t need to check back and forth while computing.

Take Example 1 to demonstrate:

1. points = [[10,16],[2,8],[1,6],[7,12]]
2. sort by the first element, then points = [[1, 6], [2, 8], [7, 12], [10, 16]]
3. find overlapping from the begining, [1, 6] and [2, 8], the overlapping is [2, 6] which is max(p1[0], p2[0]) and min(p1[1], p2[1])
4. if there is a point that p1[1] < p2[0], then there is no overlapping between these two points, so the point can be added to result list.

class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        points = sorted(points, key=lambda x: x[0])

        result = []

        current_point = points[0]

        points.pop(0)

        for point in points:
            if current_point[1] < point[0]:
                result.append(current_point)
                current_point = point
            else:
                current_point[0] = max(current_point[0], point[0])
                current_point[1] = min(current_point[1], point[1])
        
        result.append(current_point)

        return len(result)