Count and Say

Intuition

Every term is simply the run-length encoding of the previous one: count how many times a digit repeats in a row, then emit that count followed by the digit. Because term k only depends on term k-1, we can start from the base string "1" and apply this encoding step repeatedly. Building iteratively avoids deep recursion and lets us reuse a single string that we grow, transform, and replace on each round until we reach the requested position.

Algorithm walkthrough

1. 1Initialize the current term as the string "1".
2. 2Repeat the following construction n-1 times to advance from term 1 to term n.
3. 3Scan the current term, and for each group of equal adjacent digits count the run length.
4. 4Append the run length followed by the repeated digit to a fresh builder for the next term.
5. 5Replace the current term with the newly built string, and after all rounds return it.

Implementation

class Solution:
    def countAndSay(self, n: int) -> str:
        result = "1"
        for _ in range(n - 1):
            current = []
            i = 0
            while i < len(result):
                count = 1
                while i + 1 < len(result) and result[i] == result[i + 1]:
                    i += 1
                    count += 1
                current.append(str(count))
                current.append(result[i])
                i += 1
            result = "".join(current)
        return result