Leetcoder - Distinct Subsequences (DP)

 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.


public class Solution {
    public int numDistinct(String S, String T) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(S.length()<T.length()) {
            return 0;
        }
        int c[][] = new int[S.length()][T.length()];

        if(S.charAt(0)==T.charAt(0)) {
            c[0][0] = 1;
        }
        for(int i=1;i<S.length();i++) {
            c[i][0] = c[i-1][0];
            if(S.charAt(i)==T.charAt(0)) {
                c[i][0]++;
            }
        }
     
        for(int j=1;j<T.length();j++) {
            for(int i=j;i<S.length();i++) {
                if(S.charAt(i)==T.charAt(j)) {
                    c[i][j] = c[i-1][j] + c[i-1][j-1];
                } else {
                    c[i][j] = c[i-1][j];
                }
            }
        }
        return c[S.length()-1][T.length()-1];
    }
}

====================

public class Solution {
    public int numDistinct(String s, String t) {
        int len1 = s.length();
        int len2 = t.length();
        int dp[][] = new int[len1 + 1][len2 + 1];
        for (int i = 0; i <= len1; i++) {
            dp[i][0] = 1;
        }
        for (int i = 1; i <= len1; i++) {
            for (int j = 1; j <= len2; j++) {
                if (s.charAt(i - 1) == t.charAt(j - 1)) {
                    dp[i][j] = dp[i - 1][j -1] + dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i - 1][j];
                }
            }
        }
        return dp[len1][len2];
    }
}