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

## Solution 1: Simulation

Time complexity: O(max(n,m) / min(n, m))
Space complexity: O(1)

No code

## Solution 2: Simualtion + Math

For the case of 100, 3
100 – 3 = 97
97 – 3 = 94

4 – 3 = 1
Swap
3 – 1 = 2
2 – 1 = 1
1 – 1 = 0
It takes 36 steps.

We can do 100 / 3 to skip 33 steps
100 %= 3 = 1
3 / 1 = 3 to skip 3 steps
3 %= 1 = 0
total is 33 + 3 = 36.

Time complexity: O(logn) ?
Space complexity: O(1)

## C++

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