Question

452. Minimum Number of Arrows to Burst Balloons

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Description

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

Solutions

Solution 1

class Solution:
  def findMinArrowShots(self, points: List[List[int]]) -> int:
    ans, last = 0, -inf
    for a, b in sorted(points, key=lambda x: x[1]):
      if a > last:
        ans += 1
        last = b
    return ans

class Solution {
  public int findMinArrowShots(int[][] points) {
    // 直接 a[1] - b[1] 可能会溢出
    Arrays.sort(points, Comparator.comparingInt(a -> a[1]));
    int ans = 0;
    long last = -(1L << 60);
    for (var p : points) {
      int a = p[0], b = p[1];
      if (a > last) {
        ++ans;
        last = b;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int findMinArrowShots(vector<vector<int>>& points) {
    sort(points.begin(), points.end(), [](vector<int>& a, vector<int>& b) {
      return a[1] < b[1];
    });
    int ans = 0;
    long long last = -(1LL << 60);
    for (auto& p : points) {
      int a = p[0], b = p[1];
      if (a > last) {
        ++ans;
        last = b;
      }
    }
    return ans;
  }
};

func findMinArrowShots(points [][]int) (ans int) {
  sort.Slice(points, func(i, j int) bool { return points[i][1] < points[j][1] })
  last := -(1 << 60)
  for _, p := range points {
    a, b := p[0], p[1]
    if a > last {
      ans++
      last = b
    }
  }
  return
}

function findMinArrowShots(points: number[][]): number {
  points.sort((a, b) => a[1] - b[1]);
  let ans = 0;
  let last = -Infinity;
  for (const [a, b] of points) {
    if (last < a) {
      ans++;
      last = b;
    }
  }
  return ans;
}

public class Solution {
  public int FindMinArrowShots(int[][] points) {
    Array.Sort(points, (a, b) => a[1] < b[1] ? -1 : a[1] > b[1] ? 1 : 0);
    int ans = 0;
    long last = long.MinValue;
    foreach (var point in points) {
      if (point[0] > last) {
        ++ans;
        last = point[1];
      }
    }
    return ans;
  }
}