Given a string S and a string T, count the number of distinct subsequences of S which equals T.
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).
Example 1:
Input: S = "rabbbit", T = "rabbit" Output: 3 Explanation:
As shown below, there are 3 ways you can generate "rabbit" from S.
(The caret symbol ^ means the chosen letters)
rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^
Example 2:
Input: S = "babgbag", T = "bag" Output: 5 Explanation:
As shown below, there are 5 ways you can generate "bag" from S.
(The caret symbol ^ means the chosen letters)
babgbag
^^ ^
babgbag
^^ ^
babgbag
^ ^^
babgbag
^ ^^
babgbag
^^^
S = rabb
T = rabb
now the result is 1
at this time , add a 'b' after S
'b' is equal to T.back(),so that res of (rabbb,rabb) is (rabb,rab) + (rabb,rabb).
res is three
code
//this is a origin version
class Solution {
public:
int numDistinct(string s, string t) {
vector<vector<long long>> dp(s.size()+1,vector<long long>(t.size()+1));
for(int i=0;i<s.size()+1;++i)
dp[i][0]=1;
for(int i=0;i<s.size();++i){
for(int j=0;j<t.size();++j){
dp[i+1][j+1]=dp[i][j+1];
if(s[i]==t[j])
dp[i+1][j+1]+=dp[i][j];
}
}
return dp.back().back();
}
};
//this is a simplified version,the origin version is two-deminsion array.
class Solution {
public:
int numDistinct(string s, string t) {
vector<int> dp (t.size()+1,0);
dp[0]=1;
for(int i=1;i<=s.size();++i)
for(int j=t.size();j>0;--j)
if(s[i-1]==t[j-1])
dp[j]+=dp[j-1];
return dp[t.size()];
}
};