452. Minimum Number of Arrows to Burst Balloons

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.

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:

Input: points = [[1,2],[3,4],[5,6],[7,8]]
Output: 4

Example 3:

Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2

Example 4:

Input: points = [[1,2]]
Output: 1

Example 5:

Input: points = [[2,3],[2,3]]
Output: 1

Constraints:

• 0 <= points.length <= 104
• points[i].length == 2
• -231 <= xstart < xend <= 231 - 1

time = O(nlogn), space = O(logn) for sort

class Solution {
    // intuition: sort by end position first, We should shoot as to the right as possible, 
    // because since balloons are sorted, this gives you the best chance to take down more balloons. 
    // Therefore the position should always be balloon[i][1] for the ith balloon.
    
    // check how many balloons I can shoot down with one shot aiming at the ending position of the current balloon. 
    // Then skip all these balloons and start again from the next one (or the leftmost remaining one) that needs another arrow.
    
    public int findMinArrowShots(int[][] points) {
        if(points.length == 0) {
            return 0;
        }
        Arrays.sort(points, (a, b) -> a[1] - b[1]); // sort by end point
        
        int arrowPos = points[0][1];
        int arrowCnt = 1;
        for(int i = 1; i < points.length; i++) {
            if(arrowPos >= points[i][0]) {    // overlap, shot at the same time
                continue;
            }
            arrowCnt++;
            arrowPos = points[i][1];
        }
        return arrowCnt;
    }
}