Count Operations to Obtain Zero

Here’s the code that implements the logic for counting the number of operations required to make either num1 or num2 zero. The given constraints and conditions are handled in the code.

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class Solution:
    def countOperations(self, num1: int, num2: int) -> int:
        # Initialize a counter to keep track of the number of operations
        operations_count = 0

        # Continue the loop until either num1 or num2 becomes zero
        while num1 != 0 and num2 != 0:
            # If num1 is greater than or equal to num2, subtract num2 from num1
            if num1 >= num2:
                # Determine how many times num2 fits into num1
                times = num1 // num2
                # Subtract num2 from num1 that many times
                num1 -= num2 * times
                # Increment the operations count by that many times
                operations_count += times
            else: # If num2 is greater than num1, subtract num1 from num2
                # Determine how many times num1 fits into num2
                times = num2 // num1
                # Subtract num1 from num2 that many times
                num2 -= num1 * times
                # Increment the operations count by that many times
                operations_count += times

        return operations_count

The code loops until either num1 or num2 is zero, and in each iteration, it subtracts the smaller number from the larger number, updating the count accordingly. The use of the division operator // allows us to perform the subtraction multiple times in a single step if the numbers are multiples of each other, thereby optimizing the code.