452, Minimum Number of Arrows to Burst Balloons

I Problem

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

Related Topics

• Greedy
• Array
• Sorting

II Solution

Approach 1: Brute Force

Rust

/// Time Complexity: O(n^2)
/// Space Complexity: O(n)
pub fn find_min_arrow_shots(points: Vec<Vec<i32>>) -> i32 {
    points.sort_unstable_by(|a, b| a[1].cmp(&b[1]));

    let len = points.len();
    let mut burst = vec![false; len];
    let (mut idx, mut res) = (0, 0);
    let has_false = |idx: &mut usize, burst: &[bool]| {
        for i in 0..len {
            if !burst[i] {
                *idx = i;
                return true;
            }
        }
        false
    };

    while has_false(&mut idx, &burst) {
        res += 1;
        for j in idx..len {
            if points[j][0] <= points[idx][1] {
                burst[j] = true;
            }
        }
    }

    res
}

Java

BiPredicate<int[], boolean[]> hasFalse = (tup, burst) -> {
    for (int i = 0; i < burst.length; i++) {
        if (!burst[i]) {
            tup[0] = i;
            return true;
        }
    }
    return false;
};
/**
 * Time Complexity：O(n^2)
 * Space Complexity：O(n)
 */
public int findMinArrowShots(int[][] points) {
    Arrays.sort(points, Comparator.comparingInt(a -> a[1]));

    int len = points.length;
    boolean[] burst = new boolean[len];
    int[] tup = new int[2];

    while (this.hasFalse.test(tup, burst)) {
        tup[1]++;
        for (int j = tup[0], i = tup[0]; j < len; j++) {
            if (points[j][0] <= points[i][1]) {
                burst[j] = true;
            }
        }
    }

    return tup[1];
}

Go

// Time Complexity: O(n^2)
// Space Complexity: O(n)
func findMinArrowShots(points [][]int) int {
    slices.SortFunc(points, func(a, b []int) int {
        return cmp.Compare(a[1], b[1])
    })

    size := len(points)
    burst := make([]bool, size)
    idx, res := 0, 0
    hasFalse := func() bool {
        for i, v := range burst {
            if !v {
                idx = i
                return true
            }
        }
        return false
    }

    for hasFalse() {
        res++
        for j := idx; j < size; j++ {
            if points[j][0] <= points[idx][1] {
                burst[j] = true
            }
        }
    }

    return res
}

Approach 2: Greedy

Rust

/// Time Complexity: O(n*log(n))
/// Space Complexity: O(n)
pub fn find_min_arrow_shots(points: Vec<Vec<i32>>) -> i32 {
    points.sort_unstable_by(|a, b| a[1].cmp(&b[1]));

    let (mut pos, mut res) = (points[0][1], 1);
    for p in points {
        if p[0] > pos {
            res += 1;
            pos = p[1];
        }
    }

    res
}

Java

/**
 * Time Complexity：O(n*log(n))
 * Space Complexity：O(n)
 */
public int findMinArrowShots(int[][] points) {
    Arrays.sort(points, Comparator.comparingInt(a -> a[1]));

    int pos = points[0][1], res = 1;
    for (int[] p : points) {
        if (p[0] > pos) {
            res++;
            pos = p[1];
        }
    }

    return res;
}

Go

// Time Complexity: O(n*log(n))
// Space Complexity: O(n)
func findMinArrowShots(points [][]int) int {
    slices.SortFunc(points, func(a, b []int) int {
        return cmp.Compare(a[1], b[1])
    })

    pos, res := points[0][1], 1  
    for _, p := range points {
        if p[0] > pos {
            res++
            pos = p[1]
        }
    }
    
    return res
}