2169. Count Operations to Obtain Zero LeetCode Solution

In this guide, you will get 2169. Count Operations to Obtain Zero LeetCode Solution with the best time and space complexity. The solution to Count Operations to Obtain Zero problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Table of Contents

  1. Problem Statement
  2. Complexity Analysis
  3. Count Operations to Obtain Zero solution in C++
  4. Count Operations to Obtain Zero solution in Java
  5. Count Operations to Obtain Zero solution in Python
  6. Additional Resources
2169. Count Operations to Obtain Zero LeetCode Solution image

Problem Statement of Count Operations to Obtain Zero

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 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

Complexity Analysis

  • Time Complexity: O(\log(\min(\texttt{num1}, \texttt{num2})))
  • Space Complexity: O(1)

2169. Count Operations to Obtain Zero LeetCode Solution in C++

class Solution {
 public:
  int countOperations(int num1, int num2) {
    int ans = 0;

    while (num1 && num2) {
      if (num1 < num2)
        swap(num1, num2);
      ans += num1 / num2;
      num1 %= num2;
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

2169. Count Operations to Obtain Zero LeetCode Solution in Java

class Solution {
  public int countOperations(int num1, int num2) {
    int ans = 0;

    while (num1 > 0 && num2 > 0) {
      if (num1 < num2) {
        final int temp = num1;
        num1 = num2;
        num2 = temp;
      }
      ans += num1 / num2;
      num1 %= num2;
    }

    return ans;
  }
}
// code provided by PROGIEZ

2169. Count Operations to Obtain Zero LeetCode Solution in Python

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

    while num1 and num2:
      if num1 < num2:
        num1, num2 = num2, num1
      ans += num1 // num2
      num1 %= num2

    return ans
# code by PROGIEZ

Additional Resources

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