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

Approach: Greedy with Sorting by End Coordinate

Algorithm

1. Sort the balloons by their ending coordinate (xend)
2. Initialize one arrow at the end coordinate of the first balloon
3. For each subsequent balloon, if its start is greater than the current arrow position, it needs a new arrow; update the arrow position to its end
4. Return the total number of arrows

Time & Space Complexity

• Time Complexity: O(n log n) - dominated by sorting
• Space Complexity: O(n) - space used by the sorting algorithm (Java’s Arrays.sort on objects uses O(n) space)

Java Implementation

import java.util.Arrays;

public class MinimumNumberOfArrowsToBurstBalloons {

    /**
     * Return the minimum number of arrows to burst all balloons.
     * @param points Balloon intervals [xstart, xend]
     * @return Minimum number of arrows
     */
    public static int findMinArrowShots(int[][] points) {
        if (points.length == 0) {
            return 0;
        }

        // Sort by ending coordinate to avoid overflow use Integer.compare
        Arrays.sort(points, (a, b) -> Integer.compare(a[1], b[1]));

        int arrows = 1;
        int arrowPos = points[0][1];

        for (int i = 1; i < points.length; i++) {
            if (points[i][0] > arrowPos) {
                arrows++;
                arrowPos = points[i][1];
            }
        }

        return arrows;
    }

    // Test method
    public static void main(String[] args) {
        int[][] points1 = {{10, 16}, {2, 8}, {1, 6}, {7, 12}};
        System.out.println("Input: points = [[10,16],[2,8],[1,6],[7,12]]");
        System.out.println("Output: " + findMinArrowShots(points1)); // Expected: 2

        int[][] points2 = {{1, 2}, {3, 4}, {5, 6}, {7, 8}};
        System.out.println("Input: points = [[1,2],[3,4],[5,6],[7,8]]");
        System.out.println("Output: " + findMinArrowShots(points2)); // Expected: 4

        int[][] points3 = {{1, 2}, {2, 3}, {3, 4}, {4, 5}};
        System.out.println("Input: points = [[1,2],[2,3],[3,4],[4,5]]");
        System.out.println("Output: " + findMinArrowShots(points3)); // Expected: 2
    }
}

Example Walkthrough

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

1. Sort by end: [[1,6],[2,8],[7,12],[10,16]]
2. arrows=1, arrowPos=6
3. i=1 [2,8]: start 2 <= 6, no new arrow
4. i=2 [7,12]: start 7 > 6, new arrow. arrows=2, arrowPos=12
5. i=3 [10,16]: start 10 <= 12, no new arrow

Answer: 2.

Key Points

1. Greedy Choice: Always place the arrow at the earliest possible end to cover the most balloons
2. Overlapping Check: If start <= arrowPos, the balloon is burst by the current arrow
3. Overflow Safety: Use Integer.compare when sorting to avoid a - b overflow
4. Touching Intervals: A balloon touching the arrow point is burst (inclusive bounds)
5. Relation to Interval Scheduling: This is the classic greedy interval covering problem