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
.
Code:Recursive with pruning.
class Solution {
public:
int numDistinct(string s, string t) {
vector<vector<int>> mem(s.length(), vector<int>(t.length(),-1));
if(t == "") return 1;
return findNumDistinct(s,t,0, 0, mem);
}
int findNumDistinct(string &s, string &t, int ps, int pt,vector<vector<int>> &mem){
if(pt >= t.length()) return 1;
if(ps >= s.length()) return 0;
if(mem[ps][pt] != -1) return mem[ps][pt];
int res = 0;
for(int i = ps; i < s.length(); i ++){
if(s[i] == t[pt]){
res += findNumDistinct(s,t,i+1, pt+1,mem);
}
}
mem[ps][pt] = res;
return res;
}
};
Code: DP
class Solution {
public:
int numDistinct(string s, string t) {
int m = s.length();
int n = t.length();
vector<vector<int>> dp(m+1,vector<int>(n+1,0));
dp[0][0] = 1;
for(int i = 1; i <= m; i ++){
dp[i][0] = 1;
}
for(int i = 1; i <= n; i ++){
dp[0][i] = 0;
}
for(int i = 1; i <= m; i ++){
for(int j = 1; j <= n; j++){
dp[i][j] = dp[i-1][j];
if(s[i-1] == t[j-1]){
dp[i][j] += dp[i-1][j-1];
}
}
}
return dp[m][n];
}
};