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2024-03-18 Leetcode Daily Problem

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2024/03/18 12:13
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2024/03/18
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Minimum Number of Arrows to Burst Balloons - LeetCode
Can you solve this real interview question? 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
특정 풍선을 기준으로 생각해보면 해당 풍선을 지나는 화살은 반드시 하나 이상 필요하다
이 아이디어가 가장 핵심
이후 정렬 (무엇을 기준으로?) → 겹치는 구간 계산으로 아이디어 연결
정렬은 그림 그리면서 edge case 한 두개 찾으니 쉽게 감을 잡았다 → sort by (start, end)
class Solution: def findMinArrowShots(self, points: List[List[int]]) -> int: l, r = -inf, inf # overlap range arrows = 0 # result arrows for s, e in sorted(points): if r < s: # no overlap. update and reset l and r arrows, l, r = arrows + 1, s, e else: # overlap, update l to r range l, r = max(l, s), min(r, e) return arrows + 1 # + 1 for the remaining overlap
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