How do you solve Count Operations to Obtain Zero on LeetCode?

Count Operations to Obtain Zero (LeetCode #2169) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(log(min(num1, num2))) time complexity and O(1) space complexity. Key insight: Directly calculating how many times we can subtract in one operation saves time.

What is the time complexity of Count Operations to Obtain Zero?

The optimal solution for Count Operations to Obtain Zero runs in O(log(min(num1, num2))) time and uses O(1) space. The complexity is O(log(min(num1, num2))) because each operation reduces the larger number significantly, leading to a logarithmic decrease.

What is the brute force solution for Count Operations to Obtain Zero?

The brute force approach for Count Operations to Obtain Zero is Brute Force, with O(n²) time complexity and O(1) space complexity. We can simply simulate the process of subtracting the smaller number from the larger one until one of the numbers becomes zero. This is straightforward but can be inefficient for larger numbers. The optimal Optimal Solution approach improves this to O(log(min(num1, num2))).

What algorithm or data structure is used to solve Count Operations to Obtain Zero?

Count Operations to Obtain Zero uses the following concepts: Math, Simulation. The recommended patterns to study are: Mathematical Simulation, Greedy Algorithms.

What are the interview tips for Count Operations to Obtain Zero?

Always consider edge cases, such as when both numbers are equal. Think about optimizing the brute force approach before coding. Explain your thought process clearly to the interviewer.

What are common mistakes when solving Count Operations to Obtain Zero?

Forgetting to check the condition of both numbers being greater than zero. Not using integer division correctly, which can lead to incorrect counts.

#2169

Count Operations to Obtain Zero

Easy

MathSimulationMathematical SimulationGreedy Algorithms

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(log(min(num1, num2)))
Space	O(1)	O(1)

💡

Intuition

Time O(log(min(num1, num2)))Space O(1)

Instead of simulating each subtraction, we can directly calculate how many times we can subtract the smaller number from the larger one in one go. This reduces the number of operations significantly.

⚙️

Algorithm

4 steps

1Step 1: Initialize a counter for operations.
2Step 2: While both numbers are greater than zero, calculate the minimum and maximum of the two numbers.
3Step 3: Use integer division to find how many times the smaller number can be subtracted from the larger number in one operation.
4Step 4: Update the larger number and increment the operations counter accordingly.

solution.py10 lines

1def countOperations(num1, num2):
2    operations = 0
3    while num1 > 0 and num2 > 0:
4        if num1 > num2:
5            operations += num1 // num2
6            num1 %= num2
7        else:
8            operations += num2 // num1
9            num2 %= num1
10    return operations

ℹ

Complexity note: The complexity is O(log(min(num1, num2))) because each operation reduces the larger number significantly, leading to a logarithmic decrease.

1Directly calculating how many times we can subtract in one operation saves time.
2Using modulus helps in reducing the larger number efficiently.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.