Maximum Matching Of Players With Trainers - Leetcode Solution

The Maximum Matching of Players with Trainers problem can be solved using a bipartite matching algorithm. In this problem, we are given a list of players and a list of trainers, each with their own set of skills. Our task is to find the maximum number of pairs of players and trainers such that the players' skillsets match the trainers' skillsets.

The problem can be solved using the following steps:

1. Build a bipartite graph - A bipartite graph is a graph in which the nodes can be divided into two disjoint sets such that there are no edges between nodes within the same set. In this problem, we can consider the players as one set of nodes and the trainers as the other set. We can create an edge between a player and a trainer if the trainer's skillset matches the player's skillset.

2. Find the maximum matching - The maximum matching in a bipartite graph refers to the maximum number of matched pairs of nodes such that no two pairs have an edge in common. This can be found using the Hopcroft-Karp algorithm or the Ford-Fulkerson algorithm.

3. Return the maximum matching - The maximum matching of players and trainers is the maximum number of pairs of nodes that can be matched with each other without any conflicts.

Code:

Here is the Python implementation of the solution:

class Solution:
    def maxCompatibilitySum(self, students: List[List[int]], mentors: List[List[int]]) -> int:
        
        # function to check if a player matches a trainer
        def matches(player, trainer):
            return sum(a == b for a, b in zip(player, trainer))
        
        m, n = len(students), len(mentors)
        
        # build the bipartite graph
        graph = {}
        for i, player in enumerate(students):
            graph[i] = []
            for j, trainer in enumerate(mentors):
                if matches(player, trainer) == len(player):
                    graph[i].append(j)
        
        # find the maximum matching
        def dfs(node):
            if visited[node]:
                return False
            visited[node] = True
            for neighbor in graph[node]:
                if match[neighbor] is None or dfs(match[neighbor]):
                    match[neighbor] = node
                    return True
            return False
        
        match = [None] * n
        count = 0
        for player in range(m):
            visited = [False] * m
            if dfs(player):
                count += 1
        
        # return the maximum matching
        return sum(matches(students[player], mentors[match[player]]) for player in range(m))

The time complexity of this solution is O(mn^2), where m is the number of players and n is the number of trainers. This is because we are iterating over each player and trainer and checking if they match. The space complexity is also O(mn) since we are building a bipartite graph to store the matches between players and trainers.

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