Distinct Subsequences LeetCode Programming Solutions | LeetCode Problem Solutions in C++, Java, & Python [💯Correct]

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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 all Test Cases Passed, you will get a score or marks and LeetCode Coins. And after solving maximum problems, you will be getting stars. This will highlight your profile to the recruiters.

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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About LeetCode

LeetCode is one of the most well-known online judge platforms to help you enhance your skills, expand your knowledge and prepare for technical interviews. 

LeetCode is for software engineers who are looking to practice technical questions and advance their skills. Mastering the questions in each level on LeetCode is a good way to prepare for technical interviews and keep your skills sharp. They also have a repository of solutions with the reasoning behind each step.

LeetCode has over 1,900 questions for you to practice, covering many different programming concepts. Every coding problem has a classification of either EasyMedium, or Hard.

LeetCode problems focus on algorithms and data structures. Here is some topic you can find problems on LeetCode:

  • Mathematics/Basic Logical Based Questions
  • Arrays
  • Strings
  • Hash Table
  • Dynamic Programming
  • Stack & Queue
  • Trees & Graphs
  • Greedy Algorithms
  • Breadth-First Search
  • Depth-First Search
  • Sorting & Searching
  • BST (Binary Search Tree)
  • Database
  • Linked List
  • Recursion, etc.

Leetcode has a huge number of test cases and questions from interviews too like Google, Amazon, Microsoft, Facebook, Adobe, Oracle, Linkedin, Goldman Sachs, etc. LeetCode helps you in getting a job in Top MNCs. To crack FAANG Companies, LeetCode problems can help you in building your logic.

Link for the ProblemDistinct Subsequences– LeetCode Problem

Distinct Subsequences– LeetCode 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: 3
As shown below, there are 3 ways you can generate "rabbit" from S.

Example 2:

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


  • 1 <= s.length, t.length <= 1000
  • s and t consist of English letters.
Distinct Subsequences– LeetCode Solutions
Distinct Subsequences Solution in C++:
class Solution {
  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];
          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]
          dp[i][j] = dp[i - 1][j]

    return dp[m][n]

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