Leetcode problem 132 - 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

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

  let arrow = 1;
  let pos = points[0][1];

  for (let i = 1; i < points.length; i++) {
    if (pos >= points[i][0]) {
      continue;
    }

    arrow++;
    pos = points[i][1];
  }
  return arrow;
};

The time complexity of the above algorithm is O(n log n). This is because sorting the points array requires O(n log n) time, and then traversing the points array once requires O(n) time.

The space complexity is O(1). The algorithm uses only two additional variables, arrow and pos, regardless of the input size, so the space usage remains constant.