The Number of Weak Characters in the Game

class Solution:
    def numberOfWeakCharacters(self, properties: List[List[int]]) -> int:
        """
        Time Complexity: O(n * log(n))
        Space Complexity: O(1)

        Where:
            - n: size of properties.
        """
        n = len(properties)

        # Edge cases
        if n < 2:
            return 0

        # Sort by (attack, defense) tuple in ASC order
        properties.sort(key=lambda prop: (prop[0], prop[1]))

        # Select the strongest pair for the remaining question space
        max_prop = properties[-1]

        ans = 0
        next_max_prop = None
        for idx in range(n - 2, -1, -1):
            cur_prop = properties[idx]
            if cur_prop[0] == max_prop[0]:
                continue
            
            if next_max_prop is None:
                next_max_prop = cur_prop

            # then we have: cur_prop[0] < max_prop[0]
            if cur_prop[1] < max_prop[1]:
                ans += 1

            # Update the strongest pair for the remaining question space
            if idx > 0 and properties[idx - 1][0] != cur_prop[0]:
                if next_max_prop[1] > max_prop[1]:
                    max_prop = next_max_prop
                next_max_prop = None

        return ans

标签：cur,idx,1996,max,Weak,Number,prop,next,properties
来源： https://blog.csdn.net/junchen1992/article/details/120208600