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Count and Say

Number: 38

Difficulty: Medium

Paid? No

Companies: Meta, Amazon, Google, Pinterest, Microsoft, Zoho, Akuna Capital, Apple, Bloomberg, Yahoo, Wells Fargo

Problem Description

Given a positive integer n, generate the nth element of the "count-and-say" sequence—a sequence in which each element is obtained by readably describing the previous element in terms of the frequency and value of its consecutive groups of digits. The sequence starts with "1" as the first element. For instance, countAndSay(1) is "1", countAndSay(2) is "11" (one 1), countAndSay(3) is "21" (two 1s), and so on.

Key Insights

  • The problem is essentially generating a sequence where each term is produced by describing the previous term.
  • Use run-length encoding to count consecutive digits.
  • The sequence is generated iteratively from 1 up to n.
  • Iterative approach is straightforward and favored over recursion in this follow-up.

Space and Time Complexity

Time Complexity: O(n * m) where m is the average length of the sequence string generated at each step.
Space Complexity: O(m) for maintaining the current string and the next generated string.

Solution

We start with the base string "1" for n=1. For each subsequent term from 2 to n, iterate over the current string to count consecutive identical digits (run-length encoding). Append the count followed by the digit to form the next term. This iterative process continues until we reach the nth term.
Key data structures include:

  • Strings for storing the current term.
  • Temporary string builders (or string concatenation) for constructing each new term. The algorithm uses a simple loop through the string and conditionally appends to the new string when a digit changes, representing a group.

Code Solutions

# Function to generate the nth count-and-say sequence element
def countAndSay(n: int) -> str:
    # Base case: if n is 1, return "1"
    current_sequence = "1"
    
    # Iterate from 2 to n, building each sequence iteratively
    for _ in range(2, n + 1):
        next_sequence = ""  # Initialize the next sequence
        i = 0
        # Process the current sequence using run-length encoding
        while i < len(current_sequence):
            count = 1  # We have at least one occurrence of current digit
            # Count the number of same consecutive digits
            while i + 1 < len(current_sequence) and current_sequence[i] == current_sequence[i + 1]:
                count += 1
                i += 1
            # Append the count and the digit to the next sequence
            next_sequence += str(count) + current_sequence[i]
            i += 1  # Move to the next distinct digit group
        current_sequence = next_sequence  # Update current sequence for next iteration
    return current_sequence

# Example Usage:
print(countAndSay(4))  # Expected output: "1211"
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