Distinct Subsequences

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.

https://leetcode.com/problems/distinct-subsequences/

Problem statement:

The problem itself is very difficult to understand. It can be stated like this: Give a sequence S and T, how many distinct sub sequences from S equals to T? How do you define "distinct" subsequence? Clearly, the 'distinct' here mean different operation combination, not the final string of subsequence. Otherwise, the result is always 0 or 1. -- from Jason's comment

When you see string problem that is about subsequence or matching, dynamic programming method should come to mind naturally. The key is to find the initial and changing condition.

Solution:

public int numDistincts(String S, String T) {
    int[][] table = new int[S.length() + 1][T.length() + 1];

    for (int i = 0; i < S.length(); i++)
        table[i][0] = 1;

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

    return table[S.length()][T.length()];
}

Solution II: O(N) space

how to reduce to O(N) space:

http://stackoverflow.com/questions/20459262/distinct-subsequences-dp-explanation

public class Solution {

public int numDistinct(String S, String T) {
    int[] table = new int[T.length() + 1];
    table[T.length()] = 1; 

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

    return table[0]; // not return table[T.length()];   
    }
}