452. Minimum Number of Arrows to Burst Balloons.md

452. Minimum Number of Arrows to Burst Balloons

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leetcode Daily Challenge on October 10th, 2020.

leetcode-cn Daily Challenge on November 23th, 2020.

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Difficulty : Medium

Related Topics : Greedy、Sort

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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 x_start and x_end bursts by an arrow shot at x if x_start ≤ x ≤ x_end. 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] = [x_start, x_end], 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 <= 10^4
• points.length == 2
• -2^31 <= xstart < xend <= 2^31 - 1

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Solution

• mine
  • Java
    • Runtime: 14 ms, faster than 99.81%, Memory Usage: 46.4 MB, less than 17.34% of Java online submissions

      public int findMinArrowShots(int[][] p) {
          int n = p.length;
          if (n == 0) return 0;
          Arrays.sort(p, (o1, o2) -> Integer.compare(o1[1], o2[1]));
      
          int res = 1;
          int l = p[0][1];
          for (int i = 1; i < n; i++) {
              if (l < p[i][0]) {
                  res++;
                  l = p[i][1];
              }
          }
          return res;
      }
      

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