Distinct subsequence Leetcode python

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 be formed from the original strings 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  = < Code style= "Font-family:menlo,monaco,consolas, ' Courier New ', monospace; font-size:13px; PADDING:2PX 4px; Color:rgb (199,37,78); Background-color:rgb (249,242,244) ">" Rabbit "

Return 3 .

This problem is a typical DP problem. We need to maintain a DP array to find the result by sequence.

Here I had the approaches one I need to the use O (M.N) space one need to the just use O (m) space.

The time complexity is always the same. O (M.N) because we need to travesal the strings

The first method is to maintain dp[n][m]

Code is as follow

Class solution:    # @return An integer    def numdistinct (self, S, T):        dp=[[0 for J in Range (Len (T) +1)] for i in RA Nge (Len (s) +1)] for        i in range (len (s) +1):            dp[i][0]=1 for I in        Range (1,len (S) +1): For            J in Range (1,len (T) +1):                if s[i-1]==t[j-1]:                    dp[i][j]=dp[i-1][j-1]+dp[i-1][j]                else:                    Dp[i][j]=dp[i-1][j]        return DP[-1][-1]

Second method saves more space and we need to reversed the order

Class solution:    # @return An integer    def numdistinct (self, S, T):        If Len (s) ==0:            return 0        If len (T) = = 0:            return 1###         res=[0 for J in Range (Len (T) +1)]        res[0]=1        for  i in range (len (S)): for            J in Reversed (range (len (t))):                if S[I]==T[J]:                    res[j+1]=res[j]+res[j+1]        return Res[len (t)]                        

Distinct subsequence Leetcode python