Count and Say In C,CPP,JAVA,PYTHON,C#,JS

Problem Statement: Count and Say

The Count and Say sequence is a sequence of strings where each term is derived from the previous one by describing the frequency and value of the consecutive digits in the string.

For example:

• 1 is read as “one 1” → "11"
• 11 is read as “two 1s” → "21"
• 21 is read as “one 2, one 1” → "1211"
• 1211 is read as “one 1, one 2, two 1s” → "111221"
• 111221 is read as “three 1s, two 2s, one 1” → "312211"

Given an integer n, generate the nth term of the Count and Say sequence.

Input:

• An integer n (1 ≤ n ≤ 30).

Output:

• The nth term in the Count and Say sequence as a string.

Example:

Example 1:

Input: n = 1
Output: "1"

Example 2:

Input: n = 4
Output: "1211"

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Approach:

The process to generate the nth term in the Count and Say sequence involves the following steps:

1. Starting Point:
   • The first term in the sequence is always "1".
2. Iterative Process:
   • For each subsequent term, we describe the previous term by counting consecutive digits.
   • The description is done by reading the digits of the current term:
     • For a sequence of consecutive identical digits, we say how many times that digit appears, followed by the digit itself.
   • For example, if the current string is "21", we say “one 2, one 1”, which gives "1211".
3. Looping:
   • Begin with "1" as the first term and repeatedly generate the next term up to the nth term.
   • Each time, generate the next term by counting and saying the digits of the current term.
4. Time Complexity:
   • Since each term in the sequence is generated by iterating over the previous term, the time complexity is roughly proportional to the sum of the lengths of the terms up to the nth term.
   • This grows quite fast, but for n ≤ 30, this approach should be efficient enough.

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Algorithm:

1. Start with the first term "1".
2. For each i from 2 to n:
   • Generate the next term by scanning the current term and counting consecutive digits.
3. Return the nth term.

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Code Implementation:

Python:

def countAndSay(n: int) -> str:
    # The first term in the sequence is always "1"
    result = "1"
    
    # Build the sequence iteratively for n-1 times
    for _ in range(1, n):
        next_result = ""
        count = 1
        # Traverse through the current result string
        for i in range(1, len(result)):
            if result[i] == result[i - 1]:
                # If the current character is the same as the previous one, increment the count
                count += 1
            else:
                # If the current character is different, append the count and character to next_result
                next_result += str(count) + result[i - 1]
                count = 1  # Reset count for the new character
        
        # Don't forget to append the last group of characters
        next_result += str(count) + result[-1]
        
        # Update result for the next iteration
        result = next_result
    
    return result

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Explanation of Code:

1. Initialization:
   • Start with the first term "1" and assign it to result.
2. Iterating to build the sequence:
   • We loop from 1 to n-1, and in each iteration, we generate the next term by:
     • Scanning the current string.
     • Counting consecutive occurrences of each digit.
     • Appending the count followed by the digit to form the next term.
3. Updating the Result:
   • After processing the entire current term, the next term becomes the result for the next iteration.
4. Final Output:
   • Once the loop ends, result contains the nth term of the sequence, which we return.

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Time and Space Complexity:

• Time Complexity: O(m * n), where m is the length of the nth term in the sequence. For each iteration, we scan through the string to generate the next term, and since the length of the terms grows exponentially, this results in an overall time complexity of O(m * n).
• Space Complexity: O(m), where m is the length of the nth term. We only need space to store the current term and the next term.

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Example Walkthrough:

Example 1:

Input: n = 1

• The first term is "1". Since n = 1, we return "1" immediately.

Example 2:

Input: n = 4

• Start with "1".
• First iteration (n = 2): The first term is "1". We count it as “one 1”, so the next term is "11".
• Second iteration (n = 3): The term is "11". We count it as “two 1s”, so the next term is "21".
• Third iteration (n = 4): The term is "21". We count it as “one 2, one 1”, so the next term is "1211".
• The 4th term is "1211".

Example 3:

Input: n = 5

• Start with "1".
• First iteration: "1" becomes "11".
• Second iteration: "11" becomes "21".
• Third iteration: "21" becomes "1211".
• Fourth iteration: "1211" becomes "111221".
• The 5th term is "111221".

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Edge Cases:

1. n = 1: The first term is always "1", so the result is "1".
2. Larger values of n: The sequence grows quickly, but the function will work efficiently for n ≤ 30 due to the time complexity being linear with respect to the length of the terms.

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Conclusion:

The Count and Say problem can be solved efficiently using an iterative approach where each term is built from the previous one by counting and describing consecutive digits. This approach runs in linear time relative to the size of the terms, making it efficient enough for the given constraints.