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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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** Link for the Problem** – Count and Say– LeetCode Problem

Count and Say– LeetCode Problem

**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 *`n`

^{th}* term of the count-and-say sequence*.

**Example 1:**

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

**Example 2:**

Input:n = 4Output:"1211"Explanation: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"

**Constraints:**

Count and Say– LeetCode Solutions

class Solution { public: 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]) { ++count; ++i; } 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)) { ++count; ++i; } next.append(count).append(sb.charAt(i)); } 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