不同的子序列

https://leetcode.com/problems/distinct-subsequences/discuss/37327/Easy-to-understand-DP-in-Java

The idea is the following:

 

• we will build an array mem where mem[i+1][j+1] means that S[0..j] contains T[0..i] that many times as distinct subsequences. Therefor the result will be mem[T.length()][S.length()].
• we can build this array rows-by-rows:
• the first row must be filled with 1. That's because the empty string is a subsequence of any string but only 1 time. So mem[0][j] = 1 for every j. So with this we not only make our lives easier, but we also return correct value if T is an empty string.
• the first column of every rows except the first must be 0. This is because an empty string cannot contain a non-empty string as a substring -- the very first item of the array: mem[0][0] = 1, because an empty string contains the empty string 1 time.

 

So the matrix looks like this:

 

  S 0123....j
T +----------+
  |1111111111|
0 |0         |
1 |0         |
2 |0         |
. |0         |
. |0         |
i |0         |

 

From here we can easily fill the whole grid: for each (x, y), we check if S[x] == T[y] we add the previous item and the previous item in the previous row, otherwise we copy the previous item in the same row. The reason is simple:

 

• if the current character in S doesn't equal to current character T, then we have the same number of distinct subsequences as we had without the new character.
• if the current character in S equal to the current character T, then the distinct number of subsequences: the number we had before plus the distinct number of subsequences we had with less longer T and less longer S.

 

An example:
S: [acdabefbc] and T: [ab]

 

first we check with a:

 

           *  *
      S = [acdabefbc]
mem[1] = [0111222222]

 

then we check with ab:

 

               *  * ]
      S = [acdabefbc]
mem[1] = [0111222222]
mem[2] = [0000022244]

 

And the result is 4, as the distinct subsequences are:

 

      S = [a   b    ]
      S = [a      b ]
      S = [   ab    ]
      S = [   a   b ]

 

See the code in Java:

 

public int numDistinct(String S, String T) {
    // array creation
    int[][] mem = new int[T.length()+1][S.length()+1];

    // filling the first row: with 1s
    for(int j=0; j<=S.length(); j++) {
        mem[0][j] = 1;
    }
    
    // the first column is 0 by default in every other rows but the first, which we need.
    
    for(int i=0; i<T.length(); i++) {
        for(int j=0; j<S.length(); j++) {
            if(T.charAt(i) == S.charAt(j)) {
                mem[i+1][j+1] = mem[i][j] + mem[i+1][j];
            } else {
                mem[i+1][j+1] = mem[i+1][j];
            }
        }
    }
    
    return mem[T.length()][S.length()];
}