#38

Medium

auto_awesomeString-driven solution strategy

LeetCode Problem Workspace

Count and Say

Count and Say requires building the nth sequence term by iteratively applying a run-length encoding on strings of digits.

string

Problem Statement

The Count and Say sequence is a series of digit strings where each term is generated by reading aloud the previous term, describing the count of consecutive digits followed by the digit itself. Given a positive integer n, compute the nth term of this sequence efficiently using string-driven operations.

Implement a function countAndSay(n) that returns the nth string in the sequence. Start with "1" as the first term, and for each subsequent term, compress the previous string using run-length encoding (RLE) to generate the next string. Ensure your approach handles strings up to the 30th term correctly without exceeding memory limits.

Examples

Example 1

Input: See original problem statement.

Output: See original problem statement.

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

Constraints

1 <= n <= 30

Solution Approach

Iterative String Construction

Initialize the sequence with the first term '1'. Use a loop to generate each next term by iterating through the current term, counting consecutive identical digits, and appending the count and digit to build the next string.

Helper Function for Run-Length Encoding

Create a helper function that takes a string and returns its run-length encoding. This maps each consecutive digit run to a string of its count followed by the digit, reducing code duplication and clarifying the iteration logic.

Efficient Memory Handling

Construct each term in a new string to avoid modifying the previous term in place. Since string concatenation can be costly, consider using a list of characters and joining them at the end to optimize performance for larger n.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity depends on the number of digits generated for each term, which grows roughly exponentially, leading to O(2^n) in the worst case. Space complexity is dominated by the size of the string for the nth term, also O(2^n).

What Interviewers Usually Probe

Ask candidates how to generate terms iteratively instead of recursively to prevent stack overflow.
Probe understanding of run-length encoding for string sequences and handling consecutive duplicates.
Check if candidate considers memory efficiency when building strings for higher n.

Common Pitfalls or Variants

Common pitfalls

Attempting direct recursion without memoization can lead to stack overflow or repeated work.
Neglecting edge cases like n=1 or strings with only one character.
Incorrectly counting consecutive digits or misplacing the count before the digit.

Follow-up variants

Generate a custom count-and-say sequence using letters instead of digits.
Return the sequence up to nth term as a list instead of a single string.
Implement the sequence using recursion with memoization for optimization.

FAQ

What is the Count and Say sequence pattern?

It is a sequence where each term is generated by reading the previous term, counting consecutive digits, and writing the count followed by the digit.

How do I implement the run-length encoding for Count and Say?

Use a helper function that iterates through the string, counts consecutive identical digits, and appends count plus digit to build the encoded string.

What is the base case for the Count and Say sequence?

The first term n=1 is '1', which serves as the starting point for all subsequent terms.

Why is an iterative approach preferred for Count and Say?

Iterative construction avoids recursion depth issues and allows sequential building of each term efficiently.

How does the string-driven solution pattern apply here?

Each term is entirely derived from manipulating strings of digits, emphasizing the pattern of iterating, counting, and constructing new strings.

terminal

Solution

Solution 1: Simulation

The task requires outputting the appearance sequence of the $n$-th item, where the $n$-th item is the description of the $n-1$-th item in the sequence. Therefore, we iterate $n-1$ times. In each iteration, we use fast and slow pointers, denoted as j and i respectively, to record the current character's position and the position of the next character that is not equal to the current character. We then update the sequence of the previous item to be $j-i$ occurrences of the current character.

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

class Solution:
    def countAndSay(self, n: int) -> str:
        s = '1'
        for _ in range(n - 1):
            i = 0
            t = []
            while i < len(s):
                j = i
                while j < len(s) and s[j] == s[i]:
                    j += 1
                t.append(str(j - i))
                t.append(str(s[i]))
                i = j
            s = ''.join(t)
        return s

Continue Practicing

#43 multiply strings #32 longest valid parentheses #44 wildcard matching