# 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 $O(n)$.

• 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 {
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
}
}