Count and Say 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 Count and Say 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 ProblemCount and Say– LeetCode Problem

Count and Say– LeetCode Problem


The count-and-say sequence is a sequence of digit strings defined by the recursive formula:

  • countAndSay(1) = "1"
  • countAndSay(n) is the way you would “say” the digit string from countAndSay(n-1), which is then converted into a different digit string.

To determine how you “say” a digit string, split it into the minimal number of groups so that each group is a contiguous section all of the same character. Then for each group, say the number of characters, then say the character. To convert the saying into a digit string, replace the counts with a number and concatenate every saying.

For example, the saying and conversion for digit string "3322251":


Given a positive integer n, return the nth term of the count-and-say sequence.

Example 1:

Input: n = 1
Output: "1"
Explanation: This is the base case.

Example 2:

Input: n = 4
Output: "1211"
countAndSay(1) = "1"
countAndSay(2) = say "1" = one 1 = "11"
countAndSay(3) = say "11" = two 1's = "21"
countAndSay(4) = say "21" = one 2 + one 1 = "12" + "11" = "1211"


  • 1 <= n <= 30
Count and Say– LeetCode Solutions
class Solution {
  string countAndSay(int n) {
    string ans = "1";

    while (--n) {
      string next;
      for (int i = 0; i < ans.length(); ++i) {
        int count = 1;
        while (i + 1 < ans.length() && ans[i] == ans[i + 1]) {
        next += to_string(count) + ans[i];
      ans = move(next);

    return ans;
class Solution {
  public String countAndSay(int n) {
    StringBuilder sb = new StringBuilder("1");

    while (--n > 0) {
      StringBuilder next = new StringBuilder();
      for (int i = 0; i < sb.length(); ++i) {
        int count = 1;
        while (i + 1 < sb.length() && sb.charAt(i) == sb.charAt(i + 1)) {
      sb = next;

    return sb.toString();
class Solution:
  def countAndSay(self, n: int) -> str:
    ans = '1'

    for _ in range(n - 1):
      nxt = ''
      i = 0
      while i < len(ans):
        count = 1
        while i + 1 < len(ans) and ans[i] == ans[i + 1]:
          count += 1
          i += 1
        nxt += str(count) + ans[i]
        i += 1
      ans = nxt

    return ans

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