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

The optimal solution has a time complexity of O(max(num1, num2)).

What data structures are used to solve Count Operations to Obtain Zero?

This problem uses Math, Simulation. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Count Operations to Obtain Zero?

Insight 1: The problem is essentially a simulation of a subtraction process until one of the numbers becomes zero. No sophisticated mathematical formula is needed, a simple iterative approach suffices. Insight 2: A `while` loop is ideal for repeatedly performing the subtraction operation until the termination condition (either `num1` or `num2` equals 0) is met.

Which companies ask LeetCode 2169. Count Operations to Obtain Zero in interviews?

This problem is frequently asked at Accenture, Capital One, PayU.

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Count Operations to Obtain Zero - LeetCode 2169 Solution

Count Operations to Obtain Zero - Complete Solution Guide

Count Operations to Obtain Zero is LeetCode problem 2169, a Easy level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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

Detailed Explanation

The problem asks you to find the number of operations needed to reduce either `num1` or `num2` to zero. In each operation, you subtract the smaller number from the larger number. The inputs are two non-negative integers, `num1` and `num2`. The output is a single integer representing the number of operations performed. The constraints specify that both input numbers are non-negative and less than or equal to 10<sup>5</sup>.

Solution Approach

The solution uses a `while` loop to simulate the process. Inside the loop, it checks which number is larger and subtracts the smaller number from the larger one. A counter keeps track of the number of operations. The loop continues until either `num1` or `num2` becomes zero, at which point the loop terminates and the counter (number of operations) is returned.

Step-by-Step Algorithm

Step 1: Initialize a counter `count` to 0.
Step 2: Enter a `while` loop that continues as long as both `num1` and `num2` are not zero.
Step 3: Inside the loop, check if `num1` is greater than or equal to `num2`. If true, subtract `num2` from `num1`; otherwise, subtract `num1` from `num2`.
Step 4: Increment the `count` after each subtraction operation.
Step 5: After the loop finishes (one of the numbers is zero), return the value of `count`.

Key Insights

Insight 1: The problem is essentially a simulation of a subtraction process until one of the numbers becomes zero. No sophisticated mathematical formula is needed, a simple iterative approach suffices.
Insight 2: A `while` loop is ideal for repeatedly performing the subtraction operation until the termination condition (either `num1` or `num2` equals 0) is met.
Insight 3: The algorithm handles edge cases like `num1` or `num2` initially being zero gracefully. The condition `num1 != 0 && num2 != 0` ensures the loop terminates correctly.

Complexity Analysis

Time Complexity: O(max(num1, num2))

Space Complexity: O(1)

Topics

This problem involves: Math, Simulation.

Companies

Asked at: Accenture, Capital One, PayU.

Count Operations to Obtain Zero | LeetCode 2169

Easy

Math Simulation

Time: O(max(num1, num2))Space: O(1)

Solution Code

Available in:

class Solution {
    public int countOperations(int num1, int num2) {
        int count = 0;
        while (num1 != 0 && num2 != 0) {
            if (num1 >= num2) {
                num1 -= num2;
            } else {
                num2 -= num1;
            }
            count++;
        }
        return count;
    }
}

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Problem Info

Description

Problem 2169: Count Operations to Obtain Zero - Detailed explanation and solution approaches.

Topics

Math Simulation

Companies

AccentureCapital OnePayU

Stats

Difficulty:Easy

Languages:4

Companies:3

Problem #:2169

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

5 steps

1
Step 1: Initialize a counter `count` to 0.
2
Step 2: Enter a `while` loop that continues as long as both `num1` and `num2` are not zero.
3
Step 3: Inside the loop, check if `num1` is greater than or equal to `num2`. If true, subtract `num2` from `num1`; otherwise, subtract `num1` from `num2`.
4
Step 4: Increment the `count` after each subtraction operation.
5
Step 5: After the loop finishes (one of the numbers is zero), return the value of `count`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(max(num1, num2))

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the number of iterations in the `while` loop. In the worst case, the loop runs until the smaller number becomes zero. If num1 is significantly smaller than num2 (or vice-versa), the loop runs approximately `max(num1, num2) / min(num1, num2)` times which is still roughly proportional to max(num1, num2). The space complexity is O(1) because only a constant number of variables are used regardless of the input size.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Recursive Approach

A recursive function can also solve this problem. The base case is when either num1 or num2 is 0. The recursive step performs the subtraction and makes a recursive call with the updated values.

Time: O(max(num1, num2))Space: O(max(num1, num2)) due to the recursive call stack

Trade-offs: The iterative approach is generally preferred for its efficiency in space complexity. Recursion can lead to stack overflow errors for very large inputs.

GCD Approach (using Euclid's algorithm)

The number of operations is closely related to the greatest common divisor (GCD) of num1 and num2. However this approach would require additional calculation to determine exact count. It is less straight forward than the iterative solution presented.

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

Trade-offs: While GCD offers better time complexity theoretically, implementing this approach for exactly calculating the operation count requires additional steps making it less efficient than the direct iterative solution in practice for this problem.

✨

Key Insights

1Insight 1: The problem is essentially a simulation of a subtraction process until one of the numbers becomes zero. No sophisticated mathematical formula is needed, a simple iterative approach suffices.
2Insight 2: A `while` loop is ideal for repeatedly performing the subtraction operation until the termination condition (either `num1` or `num2` equals 0) is met.
3Insight 3: The algorithm handles edge cases like `num1` or `num2` initially being zero gracefully. The condition `num1 != 0 && num2 != 0` ensures the loop terminates correctly.

📚

Key Terms Explained

For Students

Edge Cases

Edge case 1: num1 = 0 or num2 = 0 initially. The while loop condition handles this correctly; the loop will not execute, and the function will return 0.
Edge case 2: num1 = num2. The loop will execute only once because after subtraction, one of them will become zero.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrect handling of the subtraction; forgetting to update either num1 or num2 after each operation.

✗Common mistake 2: Infinite loop if the termination condition isn't properly checked (both num1 and num2 becoming zero).

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Simulating processes involving repeated subtractions or resource allocation.

🏦Real-world scenario 2: Modeling certain types of mathematical games or puzzles involving reducing numbers to zero.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ Watch Out

Always test your solution with edge cases: empty input, single element, and boundary values.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: June 10, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

What should I learn from easy problems?

Easy problems introduce core patterns that appear in harder problems. Master basic operations (iteration, conditionals, simple data structures), recognize common patterns (counting, searching, basic transformations), and practice explaining your thought process clearly.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Count Operations to Obtain Zero"?

For this easy-level Math problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether simulation techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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