2169. Count Operations to Obtain Zero

Problem

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 <= 10^5

Solution (Java)

class Solution {
    public int countOperations(int num1, int num2) {
        int ans = 0;
        while ((num1 * num2) != 0) {
            if (num1 >= num2) {
                ans += num1 / num2;
                num1 = num1 % num2;
            } else {
                ans += num2 / num1;
                num2 = num2 % num1;
            }
        }
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).