What is the time complexity of Generate a String With Characters That Have Odd Counts?

The optimal solution for Generate a String With Characters That Have Odd Counts runs in O(n) time and uses O(n) space. The time complexity is O(n) because we are constructing the string in a single pass, and the space complexity is also O(n) due to the string storage.

What is the brute force solution for Generate a String With Characters That Have Odd Counts?

The brute force approach for Generate a String With Characters That Have Odd Counts is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we can generate a string of length n using any characters, ensuring that each character appears an odd number of times. However, this method can be inefficient and may require checking multiple combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Generate a String With Characters That Have Odd Counts?

Generate a String With Characters That Have Odd Counts uses the following concepts: String. The recommended patterns to study are: String Manipulation, Array Construction.

What are the interview tips for Generate a String With Characters That Have Odd Counts?

Always check if the problem can be simplified based on input properties. Think about edge cases, especially with even and odd numbers. Practice constructing strings or arrays efficiently to save time.

What are common mistakes when solving Generate a String With Characters That Have Odd Counts?

Overcomplicating the string generation process. Not considering the parity of n before constructing the string.

#1374

Generate a String With Characters That Have Odd Counts

Easy

StringString ManipulationArray Construction

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

For an optimal solution, we can directly construct the string based on whether n is odd or even. If n is odd, we can fill the string entirely with 'a's. If n is even, we can fill it with 'a's and replace the last character with 'b'. This guarantees odd counts efficiently.

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Algorithm

2 steps

1Step 1: If n is odd, return a string of 'a' repeated n times.
2Step 2: If n is even, return 'a' repeated (n-1) times followed by 'b'.

solution.py3 lines

1n = 4
2result = 'a' * n if n % 2 != 0 else 'a' * (n - 1) + 'b'
3print(result)

ℹ

Complexity note: The time complexity is O(n) because we are constructing the string in a single pass, and the space complexity is also O(n) due to the string storage.

1If n is odd, a string of 'a's is sufficient.
2If n is even, use 'a's and one 'b' to ensure odd counts.

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