LeetCode in Kotlin

2169. Count Operations to Obtain Zero

Easy

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

@Suppress("NAME_SHADOWING")
class Solution {
    fun countOperations(num1: Int, num2: Int): Int {
        var num1 = num1
        var num2 = num2
        var ans = 0
        while (num1 * num2 != 0) {
            if (num1 >= num2) {
                ans += num1 / num2
                num1 = num1 % num2
            } else {
                ans += num2 / num1
                num2 = num2 % num1
            }
        }
        return ans
    }
}

This site is open source. Improve this page.