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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 nth 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?

Solution Explanation: Count and Say

The problem asks to generate the nth string in the "count and say" sequence. The sequence is defined recursively:

  1. The first string is "1".
  2. The nth string is generated by describing the (n-1)th string. This description involves counting the consecutive occurrences of each digit and appending the count followed by the digit itself.

For example:

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

Approach: Iterative Solution

The most efficient approach is an iterative one. We start with the base case ("1") and repeatedly generate the next string in the sequence until we reach the nth string.

Algorithm:

  1. Initialization: Start with the string s = "1".
  2. Iteration: Iterate n-1 times. In each iteration:
    • Create an empty string t.
    • Iterate through the current string s.
    • For each digit:
      • Count the consecutive occurrences of that digit.
      • Append the count and the digit to the string t.
    • Update s to be t.
  3. Return: Return the final string s.

Time and Space Complexity Analysis

Time Complexity: O(n*m), where n is the input number and m is the length of the string at each iteration. The length of the string grows exponentially with n, making the overall time complexity exponential as well, although this is often masked by the relatively small input range (1 <= n <= 30). Each iteration takes roughly linear time, and the inner loop for counting consecutive numbers is also relatively fast in comparison to the overall processing.

Space Complexity: O(m) , where m is the length of the generated string. The space complexity is dominated by the size of the string being constructed at each step which, again, grows exponentially with n.

Code Implementation (Python)

def countAndSay(n: int) -> str:
    s = "1"
    for _ in range(n - 1):
        t = ""
        count = 1
        for i in range(len(s)):
            if i + 1 < len(s) and s[i] == s[i + 1]:
                count += 1
            else:
                t += str(count) + s[i]
                count = 1
        s = t
    return s

The code iterates through the string, keeping track of consecutive identical characters using the count variable. When a different character is encountered, the count and character are appended to the result string t. This process repeats until the nth string is generated.

This iterative approach provides a clear and concise way to solve the "Count and Say" problem, although it does have an exponential time and space complexity related to the growth of the string's length. For the limited input range (n <= 30), it's perfectly suitable. For substantially larger n, more sophisticated algorithms or optimizations would be necessary.