452. Minimum Number of Arrows to Burst Balloons

Problem

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 <= 10^5

• points[i].length == 2

• -2^31 <= xstart < xend <= 2^31 - 1

Solution (Java)

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;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).