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Hello **Programmers/Coders,** Today we are going to share ** solutions to the Programming problems of LeetCode Solutions in C++, Java, & Python**. At Each Problem with Successful submission with

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In this post, you will find the solution for the **Distinct Subsequences** **in C++, Java & Python-LeetCode problem**. We are providing the **correct and tested solutions** to coding problems present on **LeetCode**. If you are not able to solve any problem, then you can take help from our Blog/website.

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** Link for the Problem** – Distinct Subsequences– LeetCode Problem

Distinct Subsequences– LeetCode Problem

**Problem:**

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:3Explanation:As shown below, there are 3 ways you can generate "rabbit" from S.

brabbit

brabbit

brabbit

**Example 2:**

Input:s = "babgbag", t = "bag"Output:5Explanation:As shown below, there are 5 ways you can generate "bag" from S.

bbabagg

bgbabag

abgbbag`ba`

gbbag`babg`

bag

**Constraints:**

`1 <= s.length, t.length <= 1000`

`s`

and`t`

consist of English letters.

Distinct Subsequences– LeetCode Solutions

Distinct Subsequences Solution in C++:

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

Distinct Subsequences Solution in Java:

class Solution { public int numDistinct(String s, String t) { final int m = s.length(); final int n = t.length(); long[][] dp = new long[m + 1][n + 1]; 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.charAt(i - 1) == t.charAt(j - 1)) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j]; else dp[i][j] = dp[i - 1][j]; return (int) dp[m][n]; } }

Distinct Subsequences Solution in Python:

class Solution: def numDistinct(self, s: str, t: str) -> int: m = len(s) n = len(t) dp = [[0] * (n + 1) for _ in range(m + 1)] for i in range(m + 1): dp[i][0] = 1 for i in range(1, m + 1): for j in range(1, n + 1): if s[i - 1] == t[j - 1]: dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j] else: dp[i][j] = dp[i - 1][j] return dp[m][n]