Distinct Subsequences

• Time : $O(nm)$
• Space : $O(nm)$

int Solution::numDistinct(string A, string B) {
    int n = A.length(), m = B.length();
    vector<vector<int>> dp(n + 1, vector<int>(m + 1));
    for(int i = 0; i <= n; i++) dp[i][0] = 1;
    for(int i = 1; i <= n; i++) {
        for(int j = 1; j <= m; j++) {
            dp[i][j] = dp[i-1][j];
            if(A[i-1] == B[j-1]) dp[i][j] += dp[i-1][j-1];
        }
    }
    return dp[n][m];
}