452. Minimum Number of Arrows to Burst Balloons

Description of 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 <= 10^5
points[i].length == 2
-2^31 <= xstart < xend <= 2^31 - 1

Solution

Tags: Stack

Explanation

Consider intervals \(I_{i,j} = \lbrace x | i \le x \le j \rbrace\), \(I_{u,w} = \lbrace x | u \le x \le w \rbrace\). If they have no intersection (i.e. overlapping), nothing to do.

Otherwise, without loss of generality, let's say \( i \le u \), their intersection \( I_{i,j} \cap I_{u,w} \) becomes \(\lbrace x | \max(i,u) \le x \le \min(j,w) \rbrace\).

If \(I_1, I_2, I_3\) have an intersection (i.e. they can be burst by one arrow shot), we can get the intersection by \( (I_1 \cap I_2) \cap I_3 \).

Hence, first sort the intervals by x_start. Then keep doing intersection with last interval in the stack or push a non-overlapping interval to the stack. After that, we get the set of disjoint intervals where the number of elements equals to the minimum number of arrows that must be shot to burst all balloons.

Code (Rust)

impl Solution {
    pub fn find_min_arrow_shots(points: Vec<Vec<i32>>) -> i32 {
        let mut points = points;
        points.sort_by( |v1, v2| v1[0].cmp(&v2[0]));

        let mut stack : Vec<(i32,i32)> = vec![];

        for p in points.into_iter() {
            let (c, d) = (p[0], p[1]);
            if let Some(t) = stack.last() {
                let (a,b) = (t.0, t.1);
                if b >= c {
                    stack.pop();
                    stack.push( ( a.max(c),b.min(d) ) );
                }else{
                    stack.push( (c,d) );
                }
            }else {
                stack.push( (c,d) );
            }
        }

        return stack.len() as i32;
    }
}

Complexity

n is length of points

Time Complexity

\( T(n) = \Theta(n) \)

Auxiliary Space

\( S(n) = O(n) \)

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