452. Minimum Number of Arrows to Burst Balloons

Description

There are some spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter, and hence the x-coordinates of start and end of the diameter suffice. The start is always smaller than the end.

An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with xstart and xend bursts by an arrow shot at x if xstart ≤ x ≤ xend. There is no limit to the number of arrows that can be shot. An arrow once shot keeps traveling up infinitely.

Given an array points where points[i] = [xstart, xend], return the minimum number of arrows that must be shot to burst all balloons.

Constraints

• 0 <= points.length <= 104

• points.length == 2

• -231 <= xstart < xend <= 231 - 1

Approach

Links

• GeeksforGeeks

• Leetcode

• ProgramCreek

• YouTube

Examples

Example 1

Input: points = [[10, 16], [2, 8], [1, 6], [7, 12]]

Output: 2

Explanation: One way is to shoot one arrow for example at x = 6 (bursting the balloons [2, 8] and [1, 6]) and another arrow at x = 11 (bursting the other two balloons).

Example 2

Example 3

Example 4

Example 5

Solutions

Solution 1

/**
 * Time complexity : 
 * Space complexity : 
 */

class Solution {
    public int findMinArrowShots(int[][] points) {
        if(points == null || points.length == 0) return 0;
        
        Arrays.sort(points, (a, b) -> (a[1] > b[1])? 1: -1);
        
        int arrowCount = 1,
            arrowPos = points[0][1];
        
        for(int i = 1; i < points.length; i++) {
            if(points[i][0] > arrowPos) {
                arrowCount++;
                arrowPos = points[i][1];
            }
        }
        
        return arrowCount;
    }
}

Follow up