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;
}