LeetCode Entry

3658. GCD of Odd and Even Sums

15.07.2026 easy 2026 kotlin rust

Gcd of odds and evens

3658. GCD of Odd and Even Sums easy substack youtube

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Problem TLDR

Gcd of odds and evens

Intuition

Calculate sums, calculate gcd. Or.. return n: sum of odds is n^2, sum of evens is n(n+1).

Approach

• remember gcd as a/b, bab

Complexity

• Time complexity: \(O(n)\)

• Space complexity: \(O(1)\)

Code

    fun gcdOfOddEvenSums(n: Int) = run{
        fun gcd(a: Int, b: Int): Int = if (b==0)a else gcd(b,a%b)
        gcd((1..2*n step 2).sum(), (2..2*n step 2).sum())
    }

    pub fn gcd_of_odd_even_sums(n: i32) -> i32 { n }