Formatted question description: https://leetcode.ca/all/452.html

452. Minimum Number of Arrows to Burst Balloons (Medium)

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.length == 2
  • -231 <= xstart < xend <= 231 - 1

Related Topics:
Greedy, Sort

Similar Questions:

Solution 1. Greedy

// OJ: https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/

// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int findMinArrowShots(vector<vector<int>>& A) {
        if (A.empty()) return 0;
        sort(begin(A), end(A));
        int ans = 1, arrow = A[0][1];
        for (auto &b : A) {
            if (b[0] <= arrow) arrow = min(arrow, b[1]);
            else {
                arrow = b[1];
                ++ans;
            }
        }
        return ans;
    }
};

Or

// OJ: https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/

// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
  int findMinArrowShots(vector<vector<int>>& A) {
      sort(begin(A), end(A));
      int ans = 0, N = A.size();
      for (int i = 0; i < N; ++ans) {
          int arrow = INT_MAX;
          for (; i < N && A[i][0] <= arrow; ++i) arrow = min(arrow, A[i][1]);
      }
      return ans;
  }
}; 

Solution 2. Interval Scheduling Maximization (ISM)

// OJ: https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/

// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
    int findMinArrowShots(vector<vector<int>>& A) {
        sort(begin(A), end(A), [](auto &a, auto &b) { return a[1] < b[1]; });
        long ans = A.size(), e = LONG_MIN;
        for (auto &v : A) {
            if (v[0] <= e) --ans; // this interval overlaps with another interval. We don't need a separate arrow for it.
            else e = v[1];
        }
        return ans;
    }
};

Java

class Solution {
    public int findMinArrowShots(int[][] points) {
        if (points == null || points.length == 0)
            return 0;
        Arrays.sort(points, new Comparator<int[]>() {
            public int compare(int[] point1, int[] point2) {
                if (point1[0] != point2[0])
                    return point1[0] - point2[0];
                else
                    return point1[1] - point2[1];
            }
        });
        int arrows = 0;
        int left = points[0][0], right = points[0][1];
        int length = points.length;
        for (int i = 1; i < length; i++) {
            int[] point = points[i];
            if (point[0] > right) {
                arrows++;
                left = point[0];
                right = point[1];
            } else {
                left = Math.max(left, point[0]);
                right = Math.min(right, point[1]);
            }
        }
        arrows++;
        return arrows;
    }
}

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