# Count and Say LeetCode Solution

## Table of Contents

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”
**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`

Link: https://leetcode.com/problems/count-and-say/

## Solution #

class Solution { public: string countAndSay(int n) {

```
// base case
if(n == 1) {
return "1";
}
string result = "1";
for(int i=2;i<=n;i++) {
char previous = result\[0\];
string temp = "";
int count = 1;
for(int j=1;j<=result.length();j++) {
if(result\[j\] != previous) {
temp = temp + to\_string(count) + previous;
previous = result\[j\];
count = 1;
}else{
count++;
}
}
result = temp;
}
return result;
}
```

};