Problem statement

https://binarysearch.com/problems/Hanging-Banners/

Solution

Equal to Leetcode 0452. Minimum Number of Arrows to Burst Balloons.

Complexity

It is O(n log n) for time and space.

Code

class Solution:
    def solve(self, points):
        points.sort(key = lambda x: x[1])
        n, count = len(points), 1
        if n == 0: return 0
        curr = points[0]
        
        for i in range(n):
            if curr[1] < points[i][0]:
                count += 1
                curr = points[i]
                
        return count