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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 $O(n)$.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• class Solution {
      public int numDistinct(String s, String t) {
          int n = t.length();
          int[] f = new int[n + 1];
          f[0] = 1;
          for (char a : s.toCharArray()) {
              for (int j = n; j > 0; --j) {
                  char b = t.charAt(j - 1);
                  if (a == b) {
                      f[j] += f[j - 1];
                  }
              }
          }
          return f[n];
      }
  }
  

• class Solution {
  public:
      int numDistinct(string s, string t) {
          int n = t.size();
          unsigned long long f[n + 1];
          memset(f, 0, sizeof(f));
          f[0] = 1;
          for (char& a : s) {
              for (int j = n; j; --j) {
                  char b = t[j - 1];
                  if (a == b) {
                      f[j] += f[j - 1];
                  }
              }
          }
          return f[n];
      }
  };
  

• class Solution:
      def numDistinct(self, s: str, t: str) -> int:
          n = len(t)
          f = [1] + [0] * n
          for a in s:
              for j in range(n, 0, -1):
                  if a == t[j - 1]:
                      f[j] += f[j - 1]
          return f[n]
  
  

• func numDistinct(s string, t string) int {
  	n := len(t)
  	f := make([]int, n+1)
  	f[0] = 1
  	for _, a := range s {
  		for j := n; j > 0; j-- {
  			if b := t[j-1]; byte(a) == b {
  				f[j] += f[j-1]
  			}
  		}
  	}
  	return f[n]
  }
  

• function numDistinct(s: string, t: string): number {
      const n = t.length;
      const f: number[] = new Array(n + 1).fill(0);
      f[0] = 1;
      for (const a of s) {
          for (let j = n; j; --j) {
              const b = t[j - 1];
              if (a === b) {
                  f[j] += f[j - 1];
              }
          }
      }
      return f[n];
  }
  
  

• impl Solution {
      #[allow(dead_code)]
      pub fn num_distinct(s: String, t: String) -> i32 {
          let n = s.len();
          let m = t.len();
          let mut dp: Vec<Vec<u64>> = vec![vec![0; m + 1]; n + 1];
  
          // Initialize the dp vector
          for i in 0..=n {
              dp[i][0] = 1;
          }
  
          // Begin the actual dp process
          for i in 1..=n {
              for j in 1..=m {
                  dp[i][j] = if s.as_bytes()[i - 1] == t.as_bytes()[j - 1] {
                      dp[i - 1][j] + dp[i - 1][j - 1]
                  } else {
                      dp[i - 1][j]
                  };
              }
          }
  
          dp[n][m] as i32
      }
  }
  
  

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