[leetcode #452] Minimum Number of Arrows to Burst Balloons

Problem

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

Idea

모든 풍선을 터뜨릴 수 있는 최소한의 화살 개수를 구하는 문제다.

풍선의 영역이 겹칠 경우 하나의 화살로 여러 개의 풍선을 터뜨릴 수 있으므로 겹치는 영역을 잘 계산하면 된다.

우선 풍선의 시작점을 기준으로 오름차순으로 정렬한 뒤, 겹치는 영역을 반복해서 구한다. 겹치는 영역이 없을 경우 화살을 사용해야 하므로 화살의 개수를 하나씩 올린다. 겹치는 영역이 있을 경우 이전에 구한 겹치는 영역과 비교해 새로운 영역을 구한다.

모든 풍선을 탐색하고 난 뒤, 마지막으로 겹치는 영역에 사용할 화살이 필요하므로 지금까지 사용한 화살에 1을 더하여 리턴하면 된다.

Solution

class Solution {
    public int findMinArrowShots(int[][] points) {
        Arrays.sort(points, (int[] a, int[] b) -> a[0] - b[0]);

        int res = 0;
        int[] overlapped = new int[]{points[0][0], points[0][1]};
        for (int i=1; i < points.length; i++) {
            overlapped[0] = Math.max(overlapped[0], points[i][0]);
            overlapped[1] = Math.min(overlapped[1], points[i][1]);
            if (overlapped[0] > overlapped[1]) {
                overlapped[0] = points[i][0];
                overlapped[1] = points[i][1];
                res++;
            }
        }

        return res+1;
    }
}

Reference

https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/