Given a string S and a string T, count the number of distinct subsequences of T in S.

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of"ABCDE" while "AEC" is not).

Here is an example:
S = "rabbbit", T = "rabbit"

Return 3.

Recurse:
Judge Small: Accepted!
Judge Large: Time Limit Exceeded

int numDistinct(string S, string T) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int slen = S.length();
int tlen = T.length();
if(slen <= tlen){
if(S == T) return 1;
else return 0;
} if(S[slen-1] != T[tlen-1]) return numDistinct(S.substr(0,slen-1), T);
else
return numDistinct(S.substr(0,slen-1), T) + numDistinct(S.substr(0,slen-1), T.substr(0,tlen-1));
}

dp:
Judge Small: Accepted!
Judge Large: Accepted!

int numDistinct(string S, string T) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int col = S.length() + 1;
int row = T.length() + 1;
int** dp = new int*[row];
for(int i = 0; i < row; ++i)
dp[i] = new int[col]; for(int i = 0; i < row; ++i)
dp[i][0] = 0;
for(int j = 0; j < col; ++j)
dp[0][j] = 1; for(int i = 1; i < row; ++i)
for(int j = 1; j < col; ++j)
if(T[i-1] == S[j-1]) dp[i][j] = dp[i-1][j-1] + dp[i][j-1];
else dp[i][j] = dp[i][j-1]; int tmp = dp[row-1][col-1]; for(int i = 0; i < row; ++i)
delete[] dp[i];
delete[] dp; return tmp;
}

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