2169.count_operations_to_obtain_zero.py

"""
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.
"""

class Solution:
    def countOperations(self, num1: int, num2: int) -> int:
        count_op = 0

        # While neither of the two are 0
        while num1 != 0 and num2 != 0:
            if num1 >= num2:
                num1 -= num2

            else:
                num2 -= num1

            count_op += 1
        
        return count_op