How do you solve Count and Say on LeetCode?

Count and Say (LeetCode #38) can be solved using Brute Force or Optimal Solution (Iterative). The optimal approach is Optimal Solution (Iterative) with O(n) time complexity and O(n) space complexity. Key insight: The count-and-say sequence is built iteratively based on the previous term, which highlights the importance of understanding how sequences evolve.

What is the brute force solution for Count and Say?

The brute force approach for Count and Say is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach generates the count-and-say sequence iteratively, starting from the base case. For each step, we analyze the previous string to form the new string by counting consecutive characters. The optimal Optimal Solution (Iterative) approach improves this to O(n).

What algorithm or data structure is used to solve Count and Say?

Count and Say uses the following concepts: String. The recommended patterns to study are: Hash Map, Array, Two Pointers.

What are the interview tips for Count and Say?

When you first see this problem, clarify the definition of the count-and-say sequence and ensure you understand the run-length encoding. Communicate your approach clearly, explaining how you will build each sequence based on the previous one. Mention edge cases, such as n = 1, and discuss how you will handle them in your code.

What are common mistakes when solving Count and Say?

Students often forget to handle the last character count when building the new sequence. Some may try to use recursion without realizing the iterative nature of the problem, leading to stack overflow for larger n.

#38

Count and Say

Medium

StringHash MapArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution (Iterative)★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach uses an iterative method to generate the count-and-say sequence without excessive string manipulation. We build each sequence based on the previous one efficiently.

⚙️

Algorithm

4 steps

1Step 1: Start with the base case '1'.
2Step 2: For each number from 2 to n, create a new string by counting consecutive characters in the current string.
3Step 3: Use a StringBuilder (or equivalent) to efficiently build the new sequence.
4Step 4: Return the nth sequence after generating it iteratively.

solution.py19 lines

1# Full working Python code with comments
2
3def countAndSay(n):
4    if n == 1:
5        return '1'
6    current = '1'
7    for _ in range(2, n + 1):
8        next_sequence = ''
9        count = 1
10        for j in range(1, len(current)):
11            if current[j] == current[j - 1]:
12                count += 1
13            else:
14                next_sequence += str(count) + current[j - 1]
15                count = 1
16        next_sequence += str(count) + current[-1]  # Add the last counted character
17        current = next_sequence
18    return current
19

ℹ

Complexity note: The time complexity is O(n) because we generate n sequences, and each sequence is built in linear time relative to its length. The space complexity is O(n) due to the storage of the current sequence string.

1The count-and-say sequence is built iteratively based on the previous term, which highlights the importance of understanding how sequences evolve.
2Recognizing patterns in character counts can help optimize the generation of sequences, reducing unnecessary computations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.