Distinct Subsequences

Given two strings S and T. count the number of distinct subsequences of S that equals T.

e.g. S = "rabbbit", T = "rabit" return 3

Solution:

1. Denote that dp[i][j] is the number of distinct subsequences of S[0..i-1] that equals T[0..j-1].

2. Recursive Relation:

if S[i-1] == T[j-1]　　dp[i][j] = dp[i-1][j-1] + dp[i-1][j]

e.g.  S = abcc, T = abc. We can consider "abc" and "ab" plus the following 'c'. Or we can simply consider "abc" and "abc" without the following 'c'

if S[i-1] != T[j-1]　　dp[i][j] = dp[i-1][j]

e.g. S = abcdc, T = abd. We just consider "abcd" and "abd".

Therefore,

dp[i][j] = dp[i-1][j] + S[i-1] == T[j-1]? dp[i-1][j-1] : 0;

3.Initialization:

dp[i][0] = 1. As every String can contain a subsequence of an empty string.

Code:

public class Solution {
    public int numDistinct(String s, String t) {
        if(s.length()<t.length())
            return 0;
        if(s.length()==t.length() && s.equals(t))
            return 1;
        int[][] dp = new int[s.length()+1][t.length()+1];
        for(int i = 0; i <=s.length(); i++)
            dp[i][0]=1;
        for(int i = 0; i < s.length(); i++){
            for(int j = 0; j < t.length(); j++){
                if(s.charAt(i)==t.charAt(j)){
                    dp[i+1][j+1]=dp[i][j]+dp[i][j+1];
                }else{
                    dp[i+1][j+1]=dp[i][j+1];
                }
            }
        }
        return dp[s.length()][t.length()];
    }
}