Dynamic Programming on Distinct Subsequences

Datetime:2016-08-23 03:11:42         Topic: Dynamic Programming          Share        Original >>
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Dynamic programming (DP) is a group of very useful algorithms to solve searching problems. In many cases, it is easy to realize that a particular problem can be solved in DP, but you may spend a lot of time on finding the iterative equations. Distinct Subsequences is one such problem.

Here is the description from leetcode.

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 .

If you are sensitive enough, "two strings" and "subsequence" will guide you to DP immediately. From some previous experience like Edit Distance , you know it is a good idea to construct an m by n matrix, where m = T.length() and n = S.length() . Each element in the matrix, say matrix[i][j] , represents the number of distinct subsequences of string T[0:i+1] and S[0:j+1] . Now comes the hard part: how to fill in the matrix?

The matrix can be filled row by row, i.e. each time we add a single character in T and evaluate the number of distinct subsequences for all prefix of S . If T[i] != S[j] , since T[0:i+1] is a subsequence of S[0:j+1] (if exists), the new character T[i] should be matched before S[j] . As a result we have the equation matrix[i][j] = matrix[i][j-1] .

If T[i] == S[j] , there are two possible cases: (1) T[i] is the character that matches to S[j] , so we should throw away both T[i] and S[j] , and the number of subsequences should equal to matrix[i-1][j-1] ; or (2) T[i] is matched to a character before S[j] , where we can get the number from matrix[i][j-1] , as we have illustrated before. The total number of distinct subsequences is the sum of those two cases, so the equation should be matrix[i][j] = matrix[i-1][j-1] + matrix[i][j-1] .

To summarize, the core logic of this DP problem is shown in the following code block

if (S[j] != T[i])
matrix[i][j] = matrix[i][j-1];
matrix[i][j] = matrix[i-1][j-1] + matrix[i][j-1];

Since we only use two adjacent rows in the matrix at any time, the memory consumption can be cut down to $O(n)$ instead of $O(mn)$.

In conclusion, the difficulty of this common DP problem stands on finding the correct iterative equations. Some patterns are well-known for similar problems, such as the allocation of m by n matrix. However you will need to try hard to understand the exact meaning of a cell, and how to connect the meaning among multiple cells.


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