Algorithm
Problem Name: 452. Minimum Number of Arrows to Burst Balloons
Problem Link: https://leetcode.com/problems/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 <= 105points[i].length == 2-231 <= xstart < xend <= 231 - 1
Code Examples
#1 Code Example with Java Programming
Code -
Java Programming
class Solution {
public int findMinArrowShots(int[][] points) {
Arrays.sort(points, Comparator.comparingInt((int[] o) -> o[1]));
int count = 0;
int idx = 0;
while (idx < points.length) {
int currEnd = points[idx][1];
while (idx < points.length && points[idx][0] <= currEnd) {
idx++;
}
count++;
}
return count;
}
}
Input
Output
#2 Code Example with Javascript Programming
Code -
Javascript Programming
const findMinArrowShots = function(points) {
const sorted = points.sort((a, b) => a[0] - b[0])
let ans = 0
let lastX = null
for (let i = 0; i < sorted.length; i += 1) {
if (lastX && sorted[i][0] <= lastX) {
lastX = Math.min(sorted[i][1], lastX)
} else {
ans += 1
lastX = sorted[i][1]
}
}
return ans
}
Input
Output
#3 Code Example with Python Programming
Code -
Python Programming
class Solution:
def findMinArrowShots(self, p):
p.sort(key = lambda x: x[1])
(res, curr) = (1, p[0][1]) if p else (0, None)
for n in p:
if n[0] > curr: res, curr = res + 1, n[1]
return res
Input
Output
#4 Code Example with C# Programming
Code -
C# Programming
using System;
namespace LeetCode
{
public class _0452_MinimumNumberOfArrowsToBurstBalloons
{
public int FindMinArrowShots(int[][] points)
{
if (points.Length == 0) return 0;
Array.Sort(points, (a, b) => a[0].CompareTo(b[0]));
var result = 1;
int end = points[0][1];
foreach (var point in points)
{
if (point[0] < = end)
end = Math.Min(end, point[1]);
else
{
result++;
end = point[1];
}
}
return result;
}
}
}
Input
Output