Last updated on January 21st, 2025 at 02:08 am

Here, we see a Minimum Number of Arrows to Burst Balloons LeetCode Solution. This Leetcode problem is solved using different approaches in many programming languages, such as C++, Java, JavaScript, Python, etc.

List of all LeetCode Solution

Topics

Greedy

Companies

Microsoft

Level of Question

Medium

Minimum Number of Arrows to Burst Balloons LeetCode Solution

1. 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].

2. Coding Pattern Used in Solution

The coding pattern used in the provided code is “Merge Intervals”. This pattern is commonly used when solving problems that involve overlapping intervals, such as finding the minimum number of arrows to burst balloons, merging overlapping intervals, or finding free time slots.

3. Code Implementation in Different Languages

3.1 Minimum Number of Arrows to Burst Balloons C++

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;
    }
};

3.2 Minimum Number of Arrows to Burst Balloons 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;
    }
}

3.3 Minimum Number of Arrows to Burst Balloons 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;
};

3.4 Minimum Number of Arrows to Burst Balloons 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

4. Time and Space Complexity

	Time Complexity	Space Complexity
C++	O(n log n)	O(1)
Java	O(n log n)	O(n)
JavaScript	O(n log n)	O(n)
Python	O(n log n)	O(n)

• Time Complexity:
  • Sorting the intervals is the most expensive operation, which takes O(n log n), where n is the number of intervals.
  • The single pass through the sorted intervals to count arrows is O(n).
  • Overall time complexity: O(n log n).
• Space Complexity:
  • In C++, sorting is done in-place, so the space complexity is O(1).
  • In Java, JavaScript, and Python, sorting creates a new sorted array or list, which requires O(n) additional space.
  • Overall space complexity: O(1) for C++ and O(n) for Java, JavaScript, and Python.