Distinct Subsequences

Apply dynamic programing method. For string S and T, let DP[i][j] stands for the number of ways for T[0..i] to be subsequences of S[0..j]. We can easily give the iteration formular:

$$DP[i][j]=\begin{cases} DP[i-1][j-1]+DP[i][j-1] & S[i]=T[j]\\ DP[i][j-1] & otherwise \end{cases}$$

After compressed the space cost, we can come up with a solution like:

public class DistinctSubsequences {
    public int numDistinct(String s, String t) {
        if (t.isEmpty()) return 1;
        if (s.length() < t.length()) return 0;
        if (s.length() == t.length()) return s.equals(t) ? 1 : 0;

        char[] schar = s.toCharArray();
        char[] tchar = t.toCharArray();

        int[] dp = new int[tchar.length + 1];

        dp[0] = 1;

        for (int j = 1; j <= schar.length; j++) {
            char ch = schar[j - 1];

            for (int i = tchar.length; i > 0;i--) {
                if (tchar[i - 1] == ch) {
                    dp[i] += dp[i - 1];
                }
            }
        }

        return dp[tchar.length];
    }
}