Count and Say

Medium

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Topics:

StringsRecursion

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 <= n <= 30

Follow up: Could you solve it iteratively?

Count and Say

Medium

10 views

Topics:

StringsRecursion

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 <= n <= 30

Follow up: Could you solve it iteratively?

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected range for the input integer n? Is there a maximum value?
Should I expect any invalid inputs, such as negative values or zero, and if so, how should I handle them?
Are there any specific memory limitations that I should be aware of, considering the sequence grows with each iteration?
Could you provide an example of the sequence generation for n=4 or n=5 to confirm my understanding of the 'count and say' process?
Should the returned string only contain digits, or should I handle cases where the generated string exceeds the maximum integer representation?

Brute Force Solution

Approach

The 'Count and Say' sequence is generated by describing the previous number. The brute force approach starts with the initial number '1', and then repeatedly generates the next number in the sequence by meticulously describing the previous one. This process continues until we reach the desired term in the sequence.

Here's how the algorithm would work step-by-step:

Begin with the first term, which is always '1'.
To find the second term, carefully examine the first term ('1'). We see 'one 1', so the second term is '11'.
To find the third term, look at the second term ('11'). We see 'two 1s', so the third term is '21'.
Continue this process. For each term, read the previous term aloud, counting the number of consecutive digits and then stating the digit.
Repeat this 'counting' and 'saying' process for each subsequent term until you reach the term you need.

Code Implementation

def count_and_say(n):
    # Start with the initial number '1'
    current_term = '1'

    for _ in range(n - 1):
        next_term = ''
        count = 1

        # Iterate through the current term to generate the next term
        for i in range(len(current_term)):
            if i + 1 < len(current_term) and current_term[i] == current_term[i + 1]:
                count += 1
            else:
                # Describe the consecutive digits and append to the next term
                next_term += str(count) + current_term[i]
                count = 1

        current_term = next_term

    return current_term

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(n*m) – The outer loop iterates 'n' times, where 'n' is the desired term in the sequence. Inside the loop, we generate the next string by iterating through the previous string. The length of each string in the sequence is variable, represented by 'm', and it grows with 'n'. Therefore, the inner loop's cost is dependent on the length 'm' of the previous string. Thus, the time complexity is O(n*m) since we iterate 'n' times each taking 'm' steps on average, where 'm' is the length of the string generated at each step. The total operations approximates n * m, simplifying to O(n*m).

Space Complexity

O(N) – The algorithm iteratively builds the count and say sequence. In each iteration, it generates a new string representing the next term based on the previous term. The maximum length of any term in the sequence up to N is proportional to N. Thus, the space required to store these intermediate strings grows linearly with N, resulting in an auxiliary space complexity of O(N).

Optimal Solution

Approach

The Count and Say sequence builds upon the previous number by describing it. Instead of brute-force calculation, the optimal approach directly constructs each subsequent number based on the description of the prior one, generating the sequence iteratively.

Here's how the algorithm would work step-by-step:

Start with the first number, which is always '1'.
To get the next number, read the previous one. For example, if the previous number is '1', you'd read it as 'one 1', so the next number is '11'.
If the previous number is '11', read it as 'two 1s', so the next number becomes '21'.
If the previous number is '21', read it as 'one 2, one 1', making the next number '1211'.
Continue this process of reading the previous number and describing it to create the next number in the sequence. Essentially, group identical, consecutive digits and count how many there are and the value of the digit.
Repeat this process the specified number of times to generate the desired number in the sequence.

Code Implementation

def count_and_say(number):
    if number <= 0:
        return ""

    sequence = "1"
    for _ in range(number - 1):
        next_sequence = ""
        count = 1
        # Iterate through the current sequence to build the next sequence.
        for i in range(len(sequence)):
            if i + 1 < len(sequence) and sequence[i] == sequence[i + 1]:
                count += 1
            else:
                # Append the count and the digit to the next sequence.
                next_sequence += str(count) + sequence[i]
                count = 1
        sequence = next_sequence
        # Sequence is updated, holding calculated next iteration.

    return sequence

Try it yourself on LeetCode

Big(O) Analysis

Time Complexity

O(m * 2^n) – The countAndSay sequence is generated iteratively. For each iteration (up to n), the length of the string roughly doubles in the worst-case scenario (e.g., '1' becomes '11', '11' becomes '21', '21' becomes '1211', and so on). Therefore, the length of the string at the nth iteration is approximately 2^n. We iterate through the string of length 2^n to generate the next string, which takes O(2^n) time. Since this process is repeated n times, the overall time complexity becomes O(n * 2^n). Additionally, m represents the initial length of the string which is '1', this value is multiplied by 2^n making the answer O(m * 2^n).

Space Complexity

O(N) – The space complexity is determined by the size of the string that we construct at each step, to represent the next number in the sequence. In the worst case, the length of the string representing each number can grow proportionally with the input N, where N is the number of iterations (or the index of the desired number in the sequence). We store the previous and current string. Therefore the auxiliary space grows linearly with N, giving us O(N) space complexity.

Edge Cases

Case	How to Handle
Input n is 0	If n is 0, return an empty string or handle as an invalid input by throwing an exception since the problem specifies positive integer.
Input n is 1	If n is 1, return "1" as the first term of the sequence.
Large input n (e.g., n > 30)	The sequence grows rapidly, and very large n could lead to extremely long strings and potential performance issues, requiring efficient string manipulation.
Integer overflow during counting	Since we are counting digits, integer overflow is unlikely to be a problem given the constraints, but consider very large repeating sequences if constraints change.
Non-integer input for n	The function should explicitly check that the input 'n' is an integer, potentially truncating or throwing an error if it is a float.
Negative input for n	The problem specifies a positive integer, so the solution should throw an exception or return a predefined error string for negative n.
Memory constraints with very large n	Generating very long strings requires efficient memory management to avoid exceeding available memory, especially for larger n.
The sequence starts with numbers other than 1	Enforce the sequence to begin with '1' and any initial number will render the test case as not passing.