Problem Description

You are given an integer n. Your task is to find the GCD (greatest common divisor) of two specific sums:

1. sumOdd: The sum of the first n odd numbers

   • For example, if n = 3, the first 3 odd numbers are 1, 3, 5, so sumOdd = 1 + 3 + 5 = 9

2. sumEven: The sum of the first n even numbers

   • For example, if n = 3, the first 3 even numbers are 2, 4, 6, so sumEven = 2 + 4 + 6 = 12

Your goal is to calculate these two sums and then return their GCD.

The key insight here is recognizing the mathematical formulas for these sums:

• The sum of the first n odd numbers equals n²
• The sum of the first n even numbers equals n(n + 1)

Once we have these values, we need to find their GCD. Since n and n + 1 are consecutive integers (and therefore coprime - their GCD is 1), the GCD of n² and n(n + 1) simplifies to n.

This explains why the solution simply returns n - it's the mathematical result of finding GCD(n², n(n + 1)).

Intuition

Let's first figure out what these sums actually are. If we write out a few examples:

• First 3 odd numbers: 1, 3, 5 → sum = 9 = 3²
• First 4 odd numbers: 1, 3, 5, 7 → sum = 16 = 4²
• First 3 even numbers: 2, 4, 6 → sum = 12 = 3 × 4
• First 4 even numbers: 2, 4, 6, 8 → sum = 20 = 4 × 5

We can observe a pattern: the sum of the first n odd numbers is always n², and the sum of the first n even numbers is always n × (n + 1).

Now we need to find GCD(n², n × (n + 1)).

Let's think about what divides both numbers. Any common divisor of n² and n × (n + 1) must divide n². Since n × (n + 1) = n × n + n, any common divisor must also divide this expression.

Here's the key insight: n clearly divides both n² and n × (n + 1). But can anything larger than n divide both?

If we factor out n from both expressions:

• n² = n × n
• n × (n + 1) = n × (n + 1)

So we're really asking: what's the GCD of n and (n + 1)? Since these are consecutive integers, they share no common factors except 1. This means n and n + 1 are coprime.

Therefore, the GCD of n² and n × (n + 1) is exactly n - nothing larger can divide both, and n definitely does divide both.

Solution Approach

The implementation is remarkably elegant once we understand the mathematical relationship.

Based on our mathematical analysis:

• Sum of first n odd numbers = n²
• Sum of first n even numbers = n(n + 1)

To find the GCD of these two values, we leverage the fact that:

1. n is a common factor of both n² and n(n + 1)
2. After factoring out n, we're left with n and (n + 1)
3. Since n and n + 1 are consecutive integers, they are coprime (their GCD is 1)

Therefore, GCD(n², n(n + 1)) = n × GCD(n, n + 1) = n × 1 = n

The solution simply returns n directly:

class Solution:
    def gcdOfOddEvenSums(self, n: int) -> int:
        return n

This is a constant time O(1) solution with O(1) space complexity. We don't need to:

• Calculate the actual sums
• Implement a GCD algorithm (like Euclidean algorithm)
• Use any loops or recursion

The mathematical proof guarantees that for any positive integer n, the answer is always n itself, making this one of the most efficient solutions possible.

Example Walkthrough

Let's walk through a concrete example with n = 4 to illustrate the solution approach.

Step 1: Identify the first n odd and even numbers

• First 4 odd numbers: 1, 3, 5, 7
• First 4 even numbers: 2, 4, 6, 8

Step 2: Calculate the sums

• sumOdd = 1 + 3 + 5 + 7 = 16
• sumEven = 2 + 4 + 6 + 8 = 20

Step 3: Verify the mathematical formulas

• sumOdd = 16 = 4² ✓
• sumEven = 20 = 4 × (4 + 1) = 4 × 5 ✓

Step 4: Find GCD(16, 20) using the mathematical insight

Instead of using the Euclidean algorithm, we apply our mathematical understanding:

• 16 = 4² = 4 × 4
• 20 = 4 × 5

We can factor out 4 from both numbers:

• GCD(16, 20) = GCD(4 × 4, 4 × 5) = 4 × GCD(4, 5)

Since 4 and 5 are consecutive integers, they share no common factors except 1:

• GCD(4, 5) = 1

Therefore:

• GCD(16, 20) = 4 × 1 = 4

Step 5: Verify the result The answer is 4, which equals our input n. This confirms that for n = 4, the GCD of the sum of the first 4 odd numbers and the sum of the first 4 even numbers is indeed 4.

This example demonstrates why the solution simply returns n - the mathematical relationship guarantees this result for any positive integer input.

Solution Implementation

Time and Space Complexity

The time complexity is O(1) because the function performs a single operation - returning the input value n. This operation takes constant time regardless of the size of the input.

The space complexity is O(1) because the function does not allocate any additional memory that scales with the input. It simply returns the input parameter without creating any auxiliary data structures or variables.

Common Pitfalls

1. Misunderstanding the Problem Statement

The code calculates the sum of ALL odd and even numbers from 1 to n, but the problem asks for the sum of the FIRST n odd numbers and FIRST n even numbers.

Incorrect interpretation:

• Sum of odd numbers from 1 to n: If n=6, sums 1+3+5 = 9
• Sum of even numbers from 2 to n: If n=6, sums 2+4+6 = 12

Correct interpretation:

• First n odd numbers: If n=3, sums 1+3+5 = 9 (first 3 odd numbers)
• First n even numbers: If n=3, sums 2+4+6 = 12 (first 3 even numbers)

Solution:

def gcdOfOddEvenSums(self, n: int) -> int:
    # Calculate sum of first n odd numbers
    odd_sum = 0
    odd_num = 1
    for i in range(n):
        odd_sum += odd_num
        odd_num += 2
  
    # Calculate sum of first n even numbers
    even_sum = 0
    even_num = 2
    for i in range(n):
        even_sum += even_num
        even_num += 2
  
    # Calculate GCD
    def gcd(a, b):
        while b:
            a, b = b, a % b
        return a
  
    return gcd(odd_sum, even_sum)

2. Over-Engineering the Solution

Given the mathematical insight that the answer is always n, implementing loops and GCD algorithms is unnecessary complexity.

Optimal solution:

def gcdOfOddEvenSums(self, n: int) -> int:
    return n

3. Not Leveraging Mathematical Formulas

Even if you want to show the calculation explicitly, using loops is inefficient when direct formulas exist:

• Sum of first n odd numbers = n²
• Sum of first n even numbers = n(n+1)

Better approach with formulas:

def gcdOfOddEvenSums(self, n: int) -> int:
    odd_sum = n * n  # n²
    even_sum = n * (n + 1)  # n(n+1)
  
    def gcd(a, b):
        while b:
            a, b = b, a % b
        return a
  
    return gcd(odd_sum, even_sum)  # Will always return n

4. Edge Case Handling Confusion

The edge case handling in the original code checks if sums are 0, but given n is a positive integer and we're summing positive numbers, neither sum can be 0. This adds unnecessary code that will never execute.