452.Minimum Number of Arrows to Burst Balloons

Tags: [sort], [queue], [range]

Link: https://leetcode.com/problems/minimum-number-of-arrows-to-burst-balloons/?tab=Description

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 104balloons.

An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with xstartand xendbursts 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).

Solution: sort, queue

class Solution(object):
    def findMinArrowShots(self, points):
        """
        :type points: List[List[int]]
        :rtype: int
        """
        if not points:
            return 0

        sorted_points_by_start = sorted(points, cmp=self.compare_point_by_start)
        queue = []
        for point in sorted_points_by_start:
            if not queue:
                queue.append(point)
            else:
                if queue[-1][1] >= point[0]:
                    queue[-1][0] = max(queue[-1][0], point[0])
                    queue[-1][1] = min(queue[-1][1], point[1])
                else:
                    queue.append(point)

        return len(queue)

    @staticmethod
    def compare_point_by_start(p1, p2):
        return p1[0] - p2[0]

Revelation:

• There are two kinds of merging of the range, one is expanding the scope: [1, 6], [2, 16] => [1, 16], and the other one is narrowing the scope: [1, 6], [2, 16] => [2, 6]. To solve this problem, we need narrow down the scopes.

Note:

• Time complexity = O(nlgn), n is the number of the given points.