Count Operations to Obtain Zero - Practice Coding Problems

Count Operations to Obtain Zero - Problem

Math Simulation Easy

You're given two non-negative integers num1 and num2. Your task is to perform a series of subtraction operations until one of the numbers becomes zero.

Operation Rule:

If num1 >= num2, subtract num2 from num1
Otherwise, subtract num1 from num2

Example: Starting with num1 = 5 and num2 = 4:

Since 5 ≥ 4, subtract: num1 = 5 - 4 = 1, num2 = 4
Since 1 < 4, subtract: num1 = 1, num2 = 4 - 1 = 3
Continue until one becomes zero...

Return the total number of operations needed.

Input & Output

example_1.py — Basic Case

$ Input: num1 = 2, num2 = 3

› Output: 3

💡 Note: Operation 1: 2 < 3, so num2 = 3 - 2 = 1. Operation 2: 2 >= 1, so num1 = 2 - 1 = 1. Operation 3: 1 >= 1, so num1 = 1 - 1 = 0. Total: 3 operations.

example_2.py — Larger Numbers

$ Input: num1 = 10, num2 = 10

› Output: 1

💡 Note: Since both numbers are equal, one operation makes num1 = 10 - 10 = 0. Total: 1 operation.

example_3.py — Edge Case

$ Input: num1 = 1, num2 = 1

› Output: 1

💡 Note: Both numbers are 1, so after one subtraction: num1 = 1 - 1 = 0. Total: 1 operation.

Constraints

0 ≤ num1, num2 ≤ 10⁶
At least one of num1 or num2 will be greater than 0
The answer will fit in a 32-bit signed integer

Visualization

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Understanding the Visualization

Setup

Start with two piles: num1 stones and num2 stones

Compare

Identify which pile has more stones

Subtract

Remove smaller pile's count from larger pile

Repeat

Continue until one pile becomes empty

Optimize

Use division to skip multiple identical operations

Key Takeaway

🎯 Key Insight: This problem is essentially the Euclidean algorithm for finding GCD, but counting steps instead of finding the result!

Asked in

a Amazon 15 G Google 8 ⊞ Microsoft 6 Apple 4

The optimal solution uses division optimization similar to the Euclidean algorithm. Instead of subtracting one by one, we use integer division to count multiple operations at once, achieving O(log min(num1, num2)) time complexity instead of O(max(num1, num2)).

Common Approaches

Approach	Time	Space	Notes
✓ Simulation (Step by Step)	O(max(num1, num2))	O(1)	Simulate each subtraction operation until one number becomes zero
Optimized with Division (GCD-like)	O(log(min(num1, num2)))	O(1)	Use division to skip multiple subtraction operations at once

Simulation (Step by Step) — Algorithm Steps

Initialize operation counter to 0
While both numbers are greater than 0:
If num1 >= num2, subtract num2 from num1
Otherwise, subtract num1 from num2
Increment operation counter
Return the total operations

Visualization

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Step-by-Step Walkthrough

Initial State

Start with num1=5, num2=4, operations=0

Operation 1

5 >= 4, so num1 = 5-4 = 1, operations=1

Operation 2

1 < 4, so num2 = 4-1 = 3, operations=2

Continue

Repeat until one number becomes 0

Code -

solution.c — C

#include <stdio.h>

int countOperations(int num1, int num2) {
    int operations = 0;
    while (num1 > 0 && num2 > 0) {
        if (num1 >= num2) {
            num1 -= num2;
        } else {
            num2 -= num1;
        }
        operations++;
    }
    return operations;
}

int main() {
    int num1, num2;
    scanf("%d %d", &num1, &num2);
    printf("%d\n", countOperations(num1, num2));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(max(num1, num2))

In worst case, we subtract 1 from a large number repeatedly

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables regardless of input size

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Simulation (Step by Step) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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