LeetCode 452. Minimum Number of Arrows to Burst Balloons

Question

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

Source: https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/

Solution

// sort by end, greedy, Time Complexity O(nlog(n))
public int findMinArrowShots(int[][] points) {
    if (points == null || points.length == 0) {
        return 0;
    }
    // sort by end in ascending order
    Arrays.sort(points, new Comparator<int[]>() {
        @Override
        public int compare(int[] o1, int[] o2) {
            // in case of overflow
            if (o1[1] == o2[1]) {
                return 0;
            } else if (o1[1] > o2[1]) {
                return 1;
            } else {
                return -1;
            }
        }
    });
    int prevEnd = points[0][1];
    int numArrow = 1;
    for (int[] point : points) {
        if (point[0] > prevEnd) { // no overlap
            numArrow++;
            prevEnd = point[1];
        }
    }
    return numArrow;
}

// sort by start, greedy, Time Complexity O(nlog(n))
public int findMinArrowShots2(int[][] points) {
    if (points == null || points.length == 0) {
        return 0;
    }
    // sort by start in ascending order
    Arrays.sort(points, new Comparator<int[]>() {
        @Override
        public int compare(int[] o1, int[] o2) {
            // in case of overflow
            if (o1[0] == o2[0]) {
                return 0;
            } else if (o1[0] > o2[0]) {
                return 1;
            } else {
                return -1;
            }
        }
    });
    int minEnd = points[0][1];
    int numArrow = 1;
    for (int[] point : points) {
        if (point[0] > minEnd) { // no overlap
            numArrow++;
            minEnd = point[1];
        } else {
            minEnd = Math.min(minEnd, point[1]);
        }
    }
    return numArrow;
}