【LeetCode OJ】Distinct Subsequences

Problem Link:

http://oj.leetcode.com/problems/distinct-subsequences/

A classic problem using Dynamic Programming technique.

Let m and n be the length of the strings T and S. Let R[i][j] be the count of distinct subsequence of T[0..i] in S[0..j]. Obviously, R[i][j] = 0 for i > j.

We initialize R[0][..] first: for j = 1..n-1, if S[j] == T[0], then R[0][j] = R[0][j-1] + 1; otherwise R[0][j] = R[0][j-1].

Then we use following recursive function to update R[i][j] bottom-up (from i = 1 to m-1 and j = i to n-1):

　　R[i][j] = R[i][j-1] + R[i-1][j-1], if S[j] == T[i]

　　R[i][j] = R[i][j-1], otherwise

The python code is as follows.

class Solution:
    # @return an integer
    def numDistinct(self, S, T):
        """
        Suppose two string S[0..n-1] and T[0..m-1], with n >= m
        DP method. Let R[i][j] be the count of distinct subsequences of T[0..i] in S[0..j].
        Obviously, R[i][j] = 0, for i > j.
        Initialization: R[0][j] from j = 0 to n-1
        Recursive Function:
            R[i][j] = R[i-1][j-1] + R[i][j-1], if T[i] == T[j]
            R[i][j] = R[i][j-1], otherwise
        """
        n = len(S)
        m = len(T)
        # Special case
        if n < m:
            return 0
        # Create the 2D array R
        R = []
        for _ in xrange(m):
            R.append([0]*n)
        # Initial R[1][0..n-1]
        if T[0] == S[0]:
            R[0][0] = 1
        for j in xrange(1,n):
            if T[0] == S[j]:
                R[0][j] = R[0][j-1] + 1
            else:
                R[0][j] = R[0][j-1]
        # Update R from i = 1 to m-1, j = 0
        for i in xrange(1, m):
            for j in xrange(i, n):
                if T[i] == S[j]:
                    R[i][j] = R[i-1][j-1] + R[i][j-1]
                else:
                    R[i][j] = R[i][j-1]
        # Return R[m-1][n-1]
        return R[m-1][n-1]