38. Count and Say

LeetCode 38. 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":

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

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, "")
}