Last updated on October 5th, 2024 at 09:08 pm

Here, We see Minimum Number of Arrows to Burst Balloons LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Minimum Number of Arrows to Burst Balloons LeetCode Solution

Problem Statement

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.

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].

1. Minimum Number of Arrows to Burst Balloons LeetCode SolutionC++

class Solution {
public:
    int findMinArrowShots(vector<vector<int>>& points) {
        sort(points.begin(), points.end());
        int lastpoint = points[0][1];
        int ans = 1;
        for(auto point : points) {
            if(point[0] > lastpoint) {
                ans++;
                lastpoint = point[1];
            }
            lastpoint = min(point[1],lastpoint);
        }
        return ans;
    }
};

2. Minimum Number of Arrows to Burst Balloons LeetCode Solution Java

class Solution {
    public int findMinArrowShots(int[][] points) {
        Arrays.sort(points, (a, b) -> Integer.compare(a[1], b[1]));
        int arrows = 1;
        int prevEnd = points[0][1];
        for (int i = 1; i < points.length; ++i) {
            if (points[i][0] > prevEnd) {
                arrows++;
                prevEnd = points[i][1];
            }
        } 
        return arrows;
    }
}

3. Minimum Number of Arrows to Burst Balloons LeetCode Solution JavaScript

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

4. Minimum Number of Arrows to Burst Balloons Solution Python

class Solution(object):
    def findMinArrowShots(self, points):
        points.sort(key=lambda x: x[0])
        arrows = 1
        end = points[0][1]
        for balloon in points[1:]:
            if balloon[0] > end: 
                arrows += 1  
                end = balloon[1] 
            else:
                end = min(end, balloon[1])
        return arrows