Regular Expression II

• Time : O(nm)
• Space : O(nm)

int Solution::isInterleave(string A, string B, string C) {
    if(A.length() + B.length() != C.length()) return 0;
    int dp[155][155] = {};
    dp[0][0] = 1;
    for(int i = 0; i <= A.length(); i++) {
        for(int j = 0; j <= B.length(); j++) {
            if(i != A.length() and C[i+j] == A[i]) dp[i+1][j] |= dp[i][j];
            if(j != B.length() and C[i+j] == B[j]) dp[i][j+1] |= dp[i][j];
        }
    }
    return dp[A.length()][B.length()];
}