algoadvance

The “Count and Say” sequence is a sequence of digit strings defined by the recursive formula:

• countAndSay(1) = "1"
• countAndSay(n) is the previous term (i.e., countAndSay(n-1)) read off. This means that you count the number of digits in groups of the same digit.

For example, the first few terms of the sequence are:

• countAndSay(1) = "1"
• countAndSay(2) = "11" (one 1)
• countAndSay(3) = "21" (two 1s)
• countAndSay(4) = "1211" (one 2 followed by one 1)
• countAndSay(5) = "111221" (one 1, one 2, then two 1s)

Given an integer n, generate the n-th term of the count-and-say sequence.

Clarifying Questions

1. What is the range of the integer n we need to handle?
   • Typically, interview problems will restrict n to a reasonably small number (e.g., 1 <= n <= 30).
2. What should we return?
   • You should return a string that represents the n-th term of the count-and-say sequence.
3. Is there a maximum length we need to consider for the sequence terms?
   • Generally, this problem won’t expect very large terms; keeping space and time efficiencies reasonable for n <= 30 should suffice.

Strategy

1. Base Case: If n is 1, then we simply return "1".

2. Recursive Build: We need to build the sequence iteratively from countAndSay(1) up to countAndSay(n).
   • For each term, take the previous term and read it, counting consecutive digits and forming the new term in the format “count” + “digit”.
3. Iterate to Get Target Term:
   • Start with the base term "1".
   • For each next term, read through the current term and build the next term by counting groups of identical digits and appending “count” + “digit” to form the new term.

Code

def countAndSay(n: int) -> str:
    # Base term
    result = "1"
    
    for _ in range(n - 1):
        result = next_term(result)
        
    return result

def next_term(current_term: str) -> str:
    next_term = []
    i = 0
    length = len(current_term)
    
    while i < length:
        count = 1
        while i + 1 < length and current_term[i] == current_term[i + 1]:
            count += 1
            i += 1
        next_term.append(f"{count}{current_term[i]}")
        i += 1
    
    return "".join(next_term)

# Example usage:
print(countAndSay(5))  # Expected output: "111221"

Time Complexity

• Time Complexity: O(M * N), where M is the length of each term which grows with n, and N is the number of terms to be generated (which is n). Specifically, since the length of terms in the sequence can grow exponentially, precise complexity analysis is intricate, but an upper-bound estimate is O(2^N) for large N, due to the exponential growth of sequence lengths.
• Space Complexity: O(M), where M is the length of the current term being generated, since that’s the major space usage besides the function call stack.

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