# Count and Say

## [Count and Say](https://leetcode.com/problems/count-and-say)

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"`: ![](https://assets.leetcode.com/uploads/2020/10/23/countandsay.jpg)

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"
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:**

* `1 <= n <= 30`

## Solutions

### 🧠 Cpp

```cpp
class Solution
{
    string rle(string s)
    {
        size_t counter = 0;
        char current = s.front();
        string res;

        for(char ch : s)
        {
            if(ch == current)
                counter++;
            else
            {
                res += to_string(counter) + current;
                current = ch;
                counter = 1;
            }
        }

        //add data from last iteration at the end
        return res + to_string(counter) + current;

    }

public:
    string countAndSay(int n)
    {
        string res = "1";

        //run-length encoding
        while(--n)
            res = rle(res);

        return res;
    }
};
```


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