Problem Informations

• Problem Index: 38
• Problem Link: Count and Say
• Topics: String

Problem Description

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 $n^{th}$ 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:

• $1 \leq n \leq 30$

Follow up: Could you solve it iteratively?

Intuition

The “count-and-say” sequence is an interesting application of run-length encoding, where each term in the sequence is derived from encoding the previous term. The goal is to transform each term by counting the consecutive digits and forming a string that describes the number of occurrences followed by the digit itself. This requires understanding the pattern of conversion from one term to the next in a recursively defined sequence.

Approach

The approach to solving this problem involves a recursive function, countAndSay, that generates the sequence of strings according to the “count-and-say” rules. The base case for the recursion is straightforward: when $n=1$, the only sequence element is the string "1". For subsequent values, we use recursion to derive the $(n-1)^{th}$ term.

1. Base Case: If $n=1$, return the string "1".

2. Recursive Call: For other cases, make a recursive call to countAndSay(n-1) to get the previous term in the sequence.

3. Run-Length Encoding:

   • Once we have the $(n-1)^{th}$ term, we need to encode this term to form the $n^{th}$ term.
   • Iterate through the characters of the $(n-1)^{th}$ term while maintaining a count of consecutive identical characters.
   • For each distinct set of consecutive characters, append the count followed by the character to a new string (currentTerm).

4. Looping Over Characters:

   • Initialize a counter count to 1 for the first character.
   • Traverse the string using an index. Compare each character with the next one.
   • If they are the same, increment the counter. If not, append the counter and the character to currentTerm, reset the counter, and continue to the next distinct character.

5. Return the Result: After the loop, currentTerm will hold the encoded string, which is the $n^{th}$ term of the sequence. This is returned as the result.

Through this recursive and encoding process, each element of the sequence is constructed based on the description of the previous element, effectively capturing the essence of the “count-and-say” methodology.

Code

C++

class Solution {
public:
    // Function to generate the nth term in the count-and-say sequence
    string countAndSay(int n) {
        // Base case: the first term is "1"
        if (n == 1) return "1";
        
        // Generate the (n-1)th term recursively
        string previousTerm = countAndSay(n - 1);
        string currentTerm = "";
        
        // Iterate through the (n-1)th term to encode it
        for (int i = 0; i < previousTerm.size(); i++) {
            int count = 1;
            // Count the number of identical consecutive characters
            while (i + 1 < previousTerm.size() && previousTerm[i] == previousTerm[i + 1]) {
                count++;
                i++;
            }
            // Append count and character to the result string
            currentTerm += to_string(count) + previousTerm[i];
        }
        
        return currentTerm;
    }
};

Python

class Solution:
    # Function to generate the nth term in the count-and-say sequence
    def countAndSay(self, n: int) -> str:
        # Base case: the first term is "1"
        if n == 1:
            return "1"
        
        # Generate the (n-1)th term recursively
        previousTerm = self.countAndSay(n - 1)
        currentTerm = ""
        
        # Iterate through the (n-1)th term to encode it
        i = 0
        while i < len(previousTerm):
            count = 1
            # Count the number of identical consecutive characters
            while i + 1 < len(previousTerm) and previousTerm[i] == previousTerm[i + 1]:
                count += 1
                i += 1
            # Append count and character to the result string
            currentTerm += str(count) + previousTerm[i]
            i += 1
        
        return currentTerm

Complexity

• Time complexity: The time complexity is $O(2^n)$. This is because each call to countAndSay depends on processing all characters of the previous term. The length of the sequence approximately doubles at each step, leading to exponential growth. Hence, processing requires $O(2^{n-1}) + O(2^{n-2}) + \ldots + O(1)$ which sums up to $O(2^n)$.

• Space complexity: The space complexity is $O(2^n)$. This accounts for the space needed to store the result of the function, as well as the stack space due to recursion. Similar to the time complexity, the space required grows exponentially since each subsequent sequence can roughly double in size, leading to the space requirement of $O(2^n)$.