How do you solve GCD of Odd and Even Sums on LeetCode?

GCD of Odd and Even Sums (LeetCode #3658) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log(min(sumOdd, sumEven))) time complexity and O(1) space complexity. Key insight: The sum of the first n odd numbers is n².

What is the time complexity of GCD of Odd and Even Sums?

The optimal solution for GCD of Odd and Even Sums runs in O(log(min(sumOdd, sumEven))) time and uses O(1) space. The GCD calculation is efficient, leading to a logarithmic time complexity based on the values of sumOdd and sumEven. Space complexity remains O(1).

What is the brute force solution for GCD of Odd and Even Sums?

The brute force approach for GCD of Odd and Even Sums is Brute Force, with O(n) time complexity and O(1) space complexity. Calculate the sums of the first n odd and even numbers directly by iterating through them. This method is straightforward but inefficient for large n. The optimal Optimal Solution approach improves this to O(log(min(sumOdd, sumEven))).

What algorithm or data structure is used to solve GCD of Odd and Even Sums?

GCD of Odd and Even Sums uses the following concepts: Math, Number Theory. The recommended patterns to study are: Mathematical Induction, Number Theory.

What are the interview tips for GCD of Odd and Even Sums?

Understand the mathematical properties of numbers. Practice deriving formulas from examples. Be ready to explain your thought process clearly.

What are common mistakes when solving GCD of Odd and Even Sums?

Forgetting the formulas for sums. Confusing odd and even number sequences.

#3658

GCD of Odd and Even Sums

Easy

MathNumber TheoryMathematical InductionNumber Theory

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(log(min(sumOdd, sumEven)))
Space	O(1)	O(1)

💡

Intuition

Time O(log(min(sumOdd, sumEven)))Space O(1)

Use the mathematical formulas for the sums of odd and even numbers, which are n² and n(n + 1) respectively. This avoids unnecessary loops.

⚙️

Algorithm

3 steps

1Step 1: Calculate sumOdd as n * n.
2Step 2: Calculate sumEven as n * (n + 1).
3Step 3: Return GCD of sumOdd and sumEven.

solution.py6 lines

1import math
2
3def gcd_of_sums(n):
4    sumOdd = n * n
5    sumEven = n * (n + 1)
6    return math.gcd(sumOdd, sumEven)

ℹ

Complexity note: The GCD calculation is efficient, leading to a logarithmic time complexity based on the values of sumOdd and sumEven. Space complexity remains O(1).

1The sum of the first n odd numbers is n².
2The sum of the first n even numbers is n(n + 1).

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