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# 38. Count and Say

## LeetCode [38. Count and Say](https://leetcode-cn.com/problems/count-and-say/)

### Description

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"`:

![](/files/0KvCt9hKJc9V4lpCmQFL)

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`

### Tags

String

### Solution

The base case is return "1" when `n == 1`. We recursively "count" `countAndSay(n-1)` and return the counted result. In the count function, we use a loop to count the length of consecutive repeated character substrings in given string, and append `str(count)` and that repeated character to result string.

### Complexity

* Time complexity: $$O(mn)$$, `len(s) × n`
* Space complexity: $$O(mn)$$

### Code

```go
func countAndSay(n int) string {
	if n == 1 {
		return "1"
	}
	return count(countAndSay(n - 1))
}

func count(s string) string {
	ans := make([]string, 0, len(s)*2)
	var j int
	for i := 0; i < len(s); i = j {
		var cnt int
		for j = i; j < len(s) && s[i] == s[j]; j++ {
			cnt++
		}
		ans = append(ans, []string{strconv.Itoa(cnt), string(s[i])}...)
	}
	return strings.Join(ans, "")
}
```


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