1947. Maximum Compatibility Score Sum(DFS/DP)

This question is quite straightforward, Just need to find all combinations and calculate the score using dfs.

This is a brute force, and remember the dfs method we need to pass the value by reference otherwise it will use the extra time to copy the vector and parameter which would cause TLE.

class Solution {
public:
    int ans = 0;
    bool visited[10] = {0};
    int dp[100][100] = {0};
    int checkScore(vector<int> &s, vector<int> &m){
        int cnt = 0;
        for(int k = 0; k < s.size(); k++){
            if(s[k] == m[k]) cnt++;
        }
        return cnt;
    }
    void dfs(vector<vector<int>> & m, vector<vector<int>> & s, int idx, int score){
        if(idx == s.size()){
            ans = max(ans, score);
            return;
        }
        for(int i = 0; i < m.size(); i++){
            if(visited[i])continue;
            visited[i] = 1;
            int n = dp[idx][i];
            dfs(m, s, idx + 1, score + n);
            visited[i] = 0;
        }
    }
    int maxCompatibilitySum(vector<vector<int>>& s, vector<vector<int>>& m) {
        //brute force match all
        for(int i = 0; i < s.size(); i++){
            for(int j = 0; j < m.size(); j++){
                dp[i][j] = checkScore(s[i], m[j]);
            }
        }
        dfs(m, s, 0, 0);
        return ans;
    }
};