115. Distinct Subsequences

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
    ^^^

Idea

Such problems about subsequence, can be solved by the idea of dynamic programming.

First, the length of T must be less than S, otherwise, the answer should be zero. Then, suppose the length of S is m and the length of T is n and we have a matrix named dp in which dp[i][j] means the answer for S[0, i - 1] and T[0, j - 1]. Subsequently, we have the following transfer equations:

1. if S[i - 1] != T[j - 1], dp[i][j] = dp[i - 1][j].
2. else, dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1].

After mn times iteration, we will get the answer and it’s just dp[m][n]

Solution

class Solution {
public:
    int numDistinct(string s, string t) {
        int m = s.length(), n = t.length();
        if (n > m) {
            return 0;
        }
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, 0));
        for (int i = 0; i <= m; ++i) {
            dp[i][0] = 1;
        }
        
        for (int i = 1; i <= m; ++i) {
            for (int j = 1; j <= n; ++j) {
                if (s[i - 1] != t[j - 1]) {
                    dp[i][j] = dp[i - 1][j];
                } else {
                    dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
                }
            }
        }
        
        return dp[m][n];
    }
};

Improve

From the above codes, we can find that the space complexity is O(mn). However, in every iteration, we only use two values, dp[i - 1][j] and dp[i - 1][j - 1], in the dp matrix. And we can improve the space complexity to O(n).

class Solution {
public:
    int numDistinct(string s, string t) {
        int m = s.length(), n = t.length();

        if (n > m) {
            return 0;
        }

        vector<int> dp(n + 1, 0);
    
        dp[0] = 1;
    
        for (int i = 1; i <= m; ++i) {
            int pre = dp[0];
            for (int j = 1; j <= n; ++j) {
                int tmp = dp[j];
                if (s[i - 1] == t[j - 1]) {
                    dp[j] = dp[j] + pre;
                }
                pre = tmp;
            } 
        }
    
        return dp[n];
    }
};