115. Distinct Subsequences.md

problem description

Given a string S and a string T, count the number of distinct subsequences of S which equals T.

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).

Example 1:

Input: S = "rabbbit", T = "rabbit" Output: 3 Explanation:

As shown below, there are 3 ways you can generate "rabbit" from S.
(The caret symbol ^ means the chosen letters)

rabbbit
^^^^ ^^
rabbbit
^^ ^^^^
rabbbit
^^^ ^^^

Example 2:

Input: S = "babgbag", T = "bag" Output: 5 Explanation:

As shown below, there are 5 ways you can generate "bag" from S.
(The caret symbol ^ means the chosen letters)

babgbag
^^ ^
babgbag
^^    ^
babgbag
^    ^^
babgbag
  ^  ^^
babgbag
    ^^^

algorithm thought

做多了题目之后会有感觉，这题用动态规划解决没问题。首先就用动态规划问题想。如果两个字符串已经匹配完毕，S串加入一个新字符，如果这个字符和T串最后一个字符不匹配。没有事情发生，总匹配数依然是没加入之前的结果。如果匹配，就是之前的S串和去掉最后一个字符的T串匹配的结果加上之前匹配结果。

S = rabb 
T = rabb

now the result is 1

at this time , add a 'b' after S

'b' is equal to T.back(),so that res of (rabbb,rabb) is (rabb,rab) + (rabb,rabb).

res is three

code

//this is a origin version
class Solution {
public:
    int numDistinct(string s, string t) {
        vector<vector<long long>> dp(s.size()+1,vector<long long>(t.size()+1));
        for(int i=0;i<s.size()+1;++i)
            dp[i][0]=1;
        for(int i=0;i<s.size();++i){
            for(int j=0;j<t.size();++j){
                dp[i+1][j+1]=dp[i][j+1];
                if(s[i]==t[j])
                    dp[i+1][j+1]+=dp[i][j];
            }
        }
        return dp.back().back();
    }
};

//this is a simplified version,the origin version is two-deminsion array. 
class Solution {
public:
    int numDistinct(string s, string t) {
        vector<int> dp (t.size()+1,0);
        dp[0]=1;
        for(int i=1;i<=s.size();++i)
            for(int j=t.size();j>0;--j)
                if(s[i-1]==t[j-1])
                    dp[j]+=dp[j-1];
        return dp[t.size()];
    }
};

algorithm analysis

原始版本是二维数组版本，遍历二维数组即可，时间复杂度O(n²)。优化后的版本使用一维数组，空间复杂度降到了O(n),不仅如此，其时间也会降低