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452. Minimum Number of Arrows to Burst Balloons

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

Thinking

One arrow bursts every balloon covering that \(x\)-coordinate; we want the fewest arrows. This is piercing all intervals with the fewest points.

Sort by the right end. The current arrow sits on the last burst balloon's right end; if the next left end is past that point, shoot again and move the pin.

Pinning the rightmost point of a kept interval covers as many later balloons as possible; sorting by the right end makes that greedy choice optimal.

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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;
    }
}

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