[LeetCode] 452. Minimum Number of Arrows to Burst Balloons

There are a number of spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it's horizontal, y-coordinates don't matter and hence the x-coordinates of start and end of the diameter suffice. Start is always smaller than end. There will be at most 104 balloons.
An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with xstart and xend bursts by an arrow shot at x if xstart≤ x ≤ xend. There is no limit to the number of arrows that can be shot. An arrow once shot keeps travelling up infinitely. The problem is to find the minimum number of arrows that must be shot to burst all balloons.
Example:
Input:
[[10,16], [2,8], [1,6], [7,12]]

Output:
2

Explanation:
One way is to shoot one arrow for example at x = 6 (bursting the balloons [2,8] and [1,6]) and another arrow at x = 11 (bursting the other two balloons).

Thought process:
The idea is to sort the array based on the balloons' left ends and use a greedy approach.
After the array is sorted, we use a counter to count the minimum number of arrows needed, and an end variable to keep track of the farthest point the current arrow can shoot at while covering as many overlapping balloons as possible. 
As we iterate through the array, we check if the current balloon's left end is smaller than "end". If so, that means the current arrow can shoot it. And we update "end" to be the minimum of itself and the balloon's right end. If not, that means the current arrow has shot the maximum number of balloons. And we move onto the next arrow.
Solution:

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public class Solution {
    public int findMinArrowShots(int[][] points) {
        if (points.length == 0) {
            return 0;
        }
        
        Arrays.sort(points, new Comparator<int[]>() {
            public int compare(int[] i1, int[] i2) {
                return i1[0] - i2[0];
            } 
        });
        
        int end = points[0][1];
        int min = 1;
        for (int i = 1; i < points.length; i++) {
            if (points[i][0] <= end) {
                end = Math.min(end, points[i][1]);
            } else {
                end = points[i][1];
                min++;
            }
        }
        
        return min;
    }
}

Time complexity:

Sorting takes O(nlogn). For loop takes O(n). The overall time complexity is O(nlogn).
Location: Philadelphia, PA, USA

Comments

  1. Eman QurashiDecember 24, 2021 at 11:50 PM

    I found that solution is very popular and helpful: https://youtu.be/c_5n_qdDRDo

    ReplyDelete

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