LeetCode-in-Java

452. Minimum Number of Arrows to Burst Balloons

Medium

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 <= 105
• points[i].length == 2
• -231 <= xstart < xend <= 231 - 1

Solution

import java.util.Arrays;

public class Solution {
    /*
     * I'm glad to have come up with this solution on my own on 10/13/2021:
     * we'll have to sort the
     * balloons by its ending points, a counter case to this is below:
     * { {0, 6}, {0, 9}, {7, 8}}
     * if we sort by starting points, then it becomes:
     * {0, 6}, {0, 9}, {7, 8}
     * this way, if we shoot 9,
     * {0, 6} won't be burst however, if we sort by ending points, then it becomes:
     * {0, 6}, {7, 8}, {0, 9}, then we shoot at 6, then at 8, this gives us the result of bursting all balloons.
     */
    public int findMinArrowShots(int[][] points) {
        Arrays.sort(points, (a, b) -> Integer.compare(a[1], b[1]));
        int minArrows = 1;
        long end = points[0][1];
        for (int i = 1; i < points.length; i++) {
            if (points[i][0] > end) {
                minArrows++;
                end = points[i][1];
            }
        }
        return minArrows;
    }
}

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