# Solving Leetcode Interviews in Seconds with AI: Count Operations to Obtain Zero


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2169" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given two non-negative integers num1 and num2. In one operation, if num1 >= num2, you must subtract num2 from num1, otherwise subtract num1 from num2.  For example, if num1 = 5 and num2 = 4, subtract num2 from num1, thus obtaining num1 = 1 and num2 = 4. However, if num1 = 4 and num2 = 5, after one operation, num1 = 4 and num2 = 1.  Return the number of operations required to make either num1 = 0 or num2 = 0.   Example 1:  Input: num1 = 2, num2 = 3 Output: 3 Explanation:  - Operation 1: num1 = 2, num2 = 3. Since num1 < num2, we subtract num1 from num2 and get num1 = 2, num2 = 3 - 2 = 1. - Operation 2: num1 = 2, num2 = 1. Since num1 > num2, we subtract num2 from num1. - Operation 3: num1 = 1, num2 = 1. Since num1 == num2, we subtract num2 from num1. Now num1 = 0 and num2 = 1. Since num1 == 0, we do not need to perform any further operations. So the total number of operations required is 3.  Example 2:  Input: num1 = 10, num2 = 10 Output: 1 Explanation:  - Operation 1: num1 = 10, num2 = 10. Since num1 == num2, we subtract num2 from num1 and get num1 = 10 - 10 = 0. Now num1 = 0 and num2 = 10. Since num1 == 0, we are done. So the total number of operations required is 1.    Constraints:  0 <= num1, num2 <= 105  

	# Explanation
	Here's an efficient solution to the problem:

*   **Key Idea:** The core idea is to repeatedly subtract the smaller number from the larger until one of them becomes zero.  Using the modulo operator (%) significantly speeds up the process, as it directly calculates the remainder after repeated subtractions. The number of subtractions equals the quotient.

*   **GCD Connection:** This process is closely related to the Euclidean algorithm for finding the Greatest Common Divisor (GCD). The number of operations required is equivalent to the number of steps to find the GCD, plus the final division step.

*   **Optimization:** Instead of repeated subtractions, use the modulo operator to directly determine the new value of the larger number. The number of operations in each step is the quotient of the division.

*   **Complexity:**
    *   Runtime: O(log(min(num1, num2))) - logarithmic time complexity due to the Euclidean algorithm-like approach.
    *   Storage: O(1) - constant space complexity.

	
	# Code
	```python
	def countOperations(num1: int, num2: int) -> int:
    """
    Counts the number of operations required to make either num1 or num2 equal to 0.

    Args:
        num1: The first non-negative integer.
        num2: The second non-negative integer.

    Returns:
        The number of operations.
    """
    operations = 0
    while num1 != 0 and num2 != 0:
        if num1 >= num2:
            operations += num1 // num2
            num1 %= num2
        else:
            operations += num2 // num1
            num2 %= num1
    return operations
	```