Count And Say

Problem

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

The count-and-say sequence is a sequence of digit strings defined by the recursive formula:

• countAndSay(1) = "1"

• countAndSay(n) is the run-length encoding of countAndSay(n - 1).

Run-length encoding (RLE) is a string compression method that works by replacing consecutive identical characters (repeated 2 or more times) with the concatenation of the character and the number marking the count of the characters (length of the run). For example, to compress the string "3322251" we replace "33" with "23", replace "222" with "32", replace "5" with "15" and replace "1" with "11". Thus the compressed string becomes "23321511".

Given a positive integer n, return the n:sup:`th` element of the count-and-say sequence.

Example 1:

Input: n = 4

Output: “1211”

Explanation:

countAndSay(1) = "1"
countAndSay(2) = RLE of "1" = "11"
countAndSay(3) = RLE of "11" = "21"
countAndSay(4) = RLE of "21" = "1211"

Example 2:

Input: n = 1

Output: “1”

Explanation:

This is the base case.

Constraints:

• 1 <= n <= 30

Follow up: Could you solve it iteratively?

Pattern

String

Solution

If n == 1, return '1'. Otherwise, get the previous count-and-say string. Iterate through the digits of the previous string. If the current digit is the same as the previous digit, increment count by 1. Otherwise, append f'{count}{previous_digit}' to the count-and-say string and reset count to 1.

Code

def countAndSay(n: int) -> str:
    """Return the ``n``-th count-and-say string.
    """
    if n == 1:
        return '1'

    count_and_say = ''
    digits = countAndSay(n - 1)
    previous_digit = digits[0]
    count = 0
    for digit in digits:
        if digit == previous_digit:
            count += 1
        else:
            count_and_say += f'{count}{previous_digit}'

            previous_digit = digit
            count = 1

    count_and_say += f'{count}{previous_digit}'

    return count_and_say

Test

>>> from count_and_say__approach_1 import countAndSay
>>> countAndSay(1)
'1'
>>> countAndSay(4)
'1211'

Complexity

Time: \(O(n \cdot m)\) — n iterations, m is sequence length

Space: \(O(m)\) — stores previous sequence

count_and_say__approach_1.countAndSay(n: int) → str

Return the n-th count-and-say string.