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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.

Β 

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].

Β 

Constraints:

  • 1 <= points.length <= 105
  • points[i].length == 2
  • -231 <= xstart < xend <= 231 - 1

Solutions

Solution 1

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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) {
        // η›΄ζŽ₯ a[1] - b[1] ε―θƒ½δΌšζΊ’ε‡Ί
        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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