# C++ backtracking using visited and skip array (beating 93.1% using 9ms)

• A revised version based on @chenhaoze94's solution

``````class Solution {
public:
int dfs(vector<bool>& visited, vector<vector<int>>& skip, int cur, int idx, int m, int n){
if(idx == n) return 1;
int res = 0;

visited[cur] = true;
for(int i = 1; i <= 9; ++i){
if(!visited[i] && (!skip[cur][i] || visited[skip[cur][i]]))
res += dfs(visited, skip, i, idx + 1, m, n);
}
visited[cur] = false;
return res + (idx >= m ? 1 : 0);
}

int numberOfPatterns(int m, int n) {
vector<vector<int>>skip(10, vector<int>(10));
skip[1][3] = skip[3][1] = 2, skip[1][7] = skip[7][1] = 4;
skip[3][9] = skip[9][3] = 6, skip[7][9] = skip[9][7] = 8;
skip[1][9] = skip[9][1] = skip[3][7] = skip[7][3] = skip[2][8] = skip[8][2] = skip[4][6] = skip[6][4] = 5;

vector<bool> visited(10, false);
int res = 0;
res += dfs(visited, skip, 1, 1, m, n) * 4;
res += dfs(visited, skip, 2, 1, m, n) * 4;
res += dfs(visited, skip, 5, 1, m, n);
return res;
}
};
``````

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