115. distinct subsequence leetcode python

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本文探讨了如何计算一个字符串中另一个字符串的唯一子序列的数量，并提供了两种优化空间复杂度的方法。通过动态规划实现，文章详细解释了两种不同的空间优化策略，帮助读者理解并解决字符串匹配问题。

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.

This problem is a typical dp problem.. we need to maintain a dp array to find the result by sequence.

here I have two approaches one I need to use O(m.n) space one need to just use O(m) space.

The time complexity is always the same. O(m.n) because we need to travesal the two 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 range(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 but 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)]