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# Distinct Subsequences LeetCode Solution

## Problem – Distinct Subsequences LeetCode Solution

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

A string’s subsequence is a new string formed from the original string by deleting some (can be none) of the characters without disturbing the remaining characters’ relative positions. (i.e., `"ACE"` is a subsequence of `"ABCDE"` while `"AEC"` is not).

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.

## Distinct Subsequences LeetCode Solution in Java

``````public int numDistinct(String S, String T) {
// array creation
int[][] mem = new int[T.length()+1][S.length()+1];

// filling the first row: with 1s
for(int j=0; j<=S.length(); j++) {
mem[j] = 1;
}

// the first column is 0 by default in every other rows but the first, which we need.

for(int i=0; i<T.length(); i++) {
for(int j=0; j<S.length(); j++) {
if(T.charAt(i) == S.charAt(j)) {
mem[i+1][j+1] = mem[i][j] + mem[i+1][j];
} else {
mem[i+1][j+1] = mem[i+1][j];
}
}
}

return mem[T.length()][S.length()];
}
``````

## Distinct Subsequences LeetCode Solution in C++

``````class Solution {
public:
int numDistinct(string s, string t) {
int m = t.length(), n = s.length();
vector<int> cur(m + 1, 0);
cur = 1;
for (int j = 1; j <= n; j++) {
int pre = 1;
for (int i = 1; i <= m; i++) {
int temp = cur[i];
cur[i] = cur[i] + (t[i - 1] == s[j - 1] ? pre : 0);
pre = temp;
}
}
return cur[m];
}
};
``````

## Distinct Subsequences LeetCode Solution in Python

``````class Solution:
def numDistinct(self, s, t):
@lru_cache(None)
def dp(i, j):
if i == -1: return j == -1
if j == -1: return j == -1
return dp(i-1, j) + (s[i] == t[j]) * dp(i-1, j-1)

return dp(len(s) - 1, len(t) - 1)
``````
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