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115. Distinct Subsequences

Description

Given two strings s and t, return the number of distinct subsequences of s which equals t.

The test cases are generated so that the answer fits on a 32-bit signed integer.

 

Example 1:

Input: s = "rabbbit", t = "rabbit"
Output: 3
Explanation:
As shown below, there are 3 ways you can generate "rabbit" from s.
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.
babgbag
babgbag
babgbag
babgbag
babgbag

 

Constraints:

  • 1 <= s.length, t.length <= 1000
  • s and t consist of English letters.

Solutions

Solution 1: Dynamic Programming

We define $f[i][j]$ as the number of schemes where the first $i$ characters of string $s$ form the first $j$ characters of string $t$. Initially, $f[i][0]=1$ for all $i \in [0,m]$.

When $i > 0$, we consider the calculation of $f[i][j]$:

  • When $s[i-1] \ne t[j-1]$, we cannot select $s[i-1]$, so $f[i][j]=f[i-1][j]$;
  • Otherwise, we can select $s[i-1]$, so $f[i][j]=f[i-1][j-1]$.

Therefore, we have the following state transition equation:

$$ f[i][j]=\left{ \begin{aligned} &f[i-1][j], &s[i-1] \ne t[j-1] \ &f[i-1][j-1]+f[i-1][j], &s[i-1]=t[j-1] \end{aligned} \right. $$

The final answer is $f[m][n]$, where $m$ and $n$ are the lengths of strings $s$ and $t$ respectively.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$.

We notice that the calculation of $f[i][j]$ is only related to $f[i-1][..]$. Therefore, we can optimize the first dimension, reducing the space complexity to $O(n)$.

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class Solution:
    def numDistinct(self, s: str, t: str) -> int:
        m, n = len(s), len(t)
        f = [[0] * (n + 1) for _ in range(m + 1)]
        for i in range(m + 1):
            f[i][0] = 1
        for i, a in enumerate(s, 1):
            for j, b in enumerate(t, 1):
                f[i][j] = f[i - 1][j]
                if a == b:
                    f[i][j] += f[i - 1][j - 1]
        return f[m][n]
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class Solution {
    public int numDistinct(String s, String t) {
        int m = s.length(), n = t.length();
        int[][] f = new int[m + 1][n + 1];
        for (int i = 0; i < m + 1; ++i) {
            f[i][0] = 1;
        }
        for (int i = 1; i < m + 1; ++i) {
            for (int j = 1; j < n + 1; ++j) {
                f[i][j] = f[i - 1][j];
                if (s.charAt(i - 1) == t.charAt(j - 1)) {
                    f[i][j] += f[i - 1][j - 1];
                }
            }
        }
        return f[m][n];
    }
}
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class Solution {
public:
    int numDistinct(string s, string t) {
        int m = s.size(), n = t.size();
        unsigned long long f[m + 1][n + 1];
        memset(f, 0, sizeof(f));
        for (int i = 0; i < m + 1; ++i) {
            f[i][0] = 1;
        }
        for (int i = 1; i < m + 1; ++i) {
            for (int j = 1; j < n + 1; ++j) {
                f[i][j] = f[i - 1][j];
                if (s[i - 1] == t[j - 1]) {
                    f[i][j] += f[i - 1][j - 1];
                }
            }
        }
        return f[m][n];
    }
};
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func numDistinct(s string,