38. Count and Say

1. 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.

2. Example

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”

3. Constraints

• 1 <= n <= 30

4. Solutions

My Accepted Solution

Time complexity: $n^2$
Space complexity: n

class Solution {
public:
    string countAndSay(int n) {
        string lastString("1"), currentString("1");
        for (int i = 2; i <= n; i++) {
            lastString = move(currentString);

            for (int left = 0, right = 0; right < lastString.size(); left = right) {
                char letter = lastString[left];
                right = lastString.find_first_not_of(letter, left);
                
                // it is slower to use to_string function, I don't be sure the count will always less than 10
                // I get this trick from the solution using 0 ms
                currentString.push_back((right == string::npos ? lastString.size() : right) - left + '0');
                currentString.push_back(letter);
            }
        }

        return currentString;
    }
};