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

Solution

/**
 * @param {number[][]} points
 * @return {number}
 */
var findMinArrowShots = function(points) {
    const intervals = points.sort((a, b) => a[1] - b[1]);
    let prev = intervals[0];
    let shots = 1;

    for (let i = 1; i < intervals.length; i++) {
        const cur = intervals[i];
        if (prev[1] < cur[0]) {
            prev = cur;
            shots++;
        }
    }
    return shots;
};

We will implement a greedy solution. First, we sort points by xend. The main idea of our greedy algorithm is that we shoot the arrow at the balloon with the smallest xend. Any ballon that overlaps at this xend is discarded (since they will also be shot by that arrow). When we reach an interval points[i] that does not overlap the current xend, we must shoot a new arrow, this time at the xend of points[i]. We repeat this process and increment the number of arrows shots accordingly.