452.Minimum Number of Arrows to Burst Balloons

tags: Medium,Array,Greedy,Sorting

452. Minimum Number of Arrows to Burst Balloons

題目描述

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] = [

$x_{start}$,

$x_{end}$] denotes a balloon whose horizontal diameter stretches between

$x_{start}$ and

$x_{end}$. 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

$x_{start}$ and

$x_{end}$ is burst by an arrow shot at x if

$x_{start}$ <= x <=

$x_{end}$. 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.

範例

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 <=
  $x_{start}$ <
  $x_{end}$ <= 2³¹ - 1

解答

Javascript

function findMinArrowShots(points) {
  points.sort((a, b) => a[0] - b[0]);
  let count = 1;
  let end = points[0][1];
  for (let i = 1; i < points.length; i++) {
    if (points[i][0] > end) {
      count++;
      end = points[i][1];
    } else {
      end = Math.min(end, points[i][1]);
    }
  }

  return count;
}

這題跟merge intervals好像
Marsgoat Jan 5, 2023

Python

class Solution:
    def findMinArrowShots(self, points: List[List[int]]) -> int:
        bolloons = sorted(points, key=lambda x: x[1])
        
        first_balloon_end = bolloons[0][1]
        arrow = 1

        for balloon_start, ballon_end in bolloons:
            if balloon_start > first_balloon_end:
                arrow += 1
                first_balloon_end = ballon_end

        return arrow

Ron Chen Jan 6, 2023

Reference

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