2169. Count Operations to Obtain Zero

EasyMathSimulation

Problem Description

You are provided with two non-negative integers num1 and num2. Your task is to perform a series of operations to reduce either num1 or num2 to zero. In a single operation, you compare num1 and num2. If num1 is greater than or equal to num2, you subtract num2 from num1. Otherwise, you subtract num1 from num2. The operation is repeated until one of the numbers becomes zero. The goal is to determine the number of operations required to achieve this.

Intuition

The key to solving the problem is recognizing that in each operation, the larger number is being reduced by the smaller number. This is reminiscent of the Euclidean algorithm, which is used to find the greatest common divisor (GCD) of two numbers, although in this problem, we're not necessarily finding the GCD, but rather bringing one of the numbers down to zero.

To arrive at the solution approach, we can use a while loop that runs as long as neither num1 nor num2 is zero. On each iteration, the operation is performed as per the described rules: if num1 is greater than or equal to num2, then num1 is to be subtracted by num2, but to make the calculation easier and more efficient, we can swap num1 and num2 instead, and then proceed to subtract num1 from num2. Each time an operation is performed, we increment a counter variable to keep track of how many operations have been carried out. We continue this process until one of the numbers is reduced to zero. At that point, we return the counter variable, which gives us the total number of operations done to reach the objective.

Learn more about Math patterns.

Solution Approach

The solution implements a straightforward iterative approach without the need for any complex algorithms, auxiliary data structures, or design patterns. Here’s a step-by-step explanation of how the solution code works:

Initialize a counter variable ans to zero. This will keep track of the number of operations performed.
Use a while loop that will continue as long as both num1 and num2 are non-zero. This loop will break when one of the numbers becomes zero, which is our stopping condition.
Inside the loop, we check if num1 is greater than or equal to num2. The objective is to always subtract the smaller number from the larger one. To ensure this, we swap num1 and num2 whenever num1 is larger, leveraging Python’s multiple assignment capability: num1, num2 = num2, num1. This keeps the invariant that num1 should always be less than or equal to num2.
Perform the subtraction operation: num2 -= num1. This effectively reduces the larger number by the value of the smaller number, mimicking one 'operation' as described in the problem.
Increment the operation counter ans by one, signifying that an operation has been completed.
Continue the loop until one of num1 or num2 reaches zero. At that point, exit the while loop.
Finally, return the counter ans, which now contains the total number of operations performed to reach the goal.

This solution is efficient because it continuously reduces the larger number, effectively halving the problem size with many of the operations, and it does so in-place without any need for additional memory. Additionally, it takes advantage of the Python multiple assignment feature for a clean and concise implementation.

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Example Walkthrough

Let's illustrate the solution approach using a small example with num1 = 7 and num2 = 4.

We initialize our counter ans to 0. This will keep track of how many operations we perform.
Our while condition is while num1 != 0 and num2 != 0, and since both num1 and num2 are non-zero, we enter the loop.
We compare num1 and num2. Since num1 (7) is greater than num2 (4), according to our rules, we need to subtract num2 from num1. However, for simplicity, we swap num1 and num2 instead, so now num1 becomes 4 and num2 becomes 7.
Then we perform the subtraction: num2 -= num1, which means num2 is now 7 - 4 = 3.
We increment ans by one, so ans = 1.
We repeat this process, now num1 (4) is still greater than num2 (3), so we swap again. Now num1 is 3 and num2 is 4.
We perform the operation num2 -= num1, so num2 is now 4 - 3 = 1.
We increment ans by one again, so ans = 2.
The process repeats with num1 being 3 and num2 being 1, no need to swap this time. After subtraction, num1 becomes 3 - 1 = 2 and ans increments to 3.
Continuing, num1 (2) is greater, so we swap. num1 is 1, num2 is 2, and after subtraction, num2 becomes 1. ans increments to 4.
Finally, with num1 being 1 and num2 also 1, we subtract and num2 becomes 0. ans increments to 5.
Now num2 is 0, the while condition breaks, and we exit the loop.
We return our counter ans, which is now 5, indicating we've performed 5 operations to reduce num2 to zero following the given rules.

Therefore, it takes 5 operations to reduce either num1 or num2 to zero, given the initial values of num1 = 7 and num2 = 4.

Solution Implementation

1class Solution:
2    def count_operations(self, num1: int, num2: int) -> int:
3        # Initialize operation count as 0
4        operation_count = 0
5      
6        # Loop until either of the numbers becomes 0
7        while num1 and num2: 
8            # Ensure num1 is the smaller number by swapping if necessary
9            if num1 >= num2:
10                num1, num2 = num2, num1
11          
12            # Subtract the smaller number (num1) from the larger number (num2)
13            num2 -= num1
14          
15            # Increment the operation count after each subtraction
16            operation_count += 1
17      
18        # Return the total count of operations performed
19        return operation_count
20

1class Solution {
2    // Counts the number of operations to make either num1 or num2 equal to 0
3    // by repeatedly subtracting the smaller value from the larger one.
4    public int countOperations(int num1, int num2) {
5        int operationsCount = 0; // Initialize the count of operations to 0
6      
7        // Loop continues as long as neither num1 nor num2 is equal to 0
8        while (num1 != 0 && num2 != 0) {
9            // If num1 is greater than or equal to num2
10            if (num1 >= num2) {
11                num1 -= num2; // Subtract num2 from num1
12            } else { // If num2 is greater than num1
13                num2 -= num1; // Subtract num1 from num2
14            }
15            operationsCount++; // Increment the count of operations
16        }
17        return operationsCount; // Return the total number of operations performed
18    }
19}
20

1class Solution {
2public:
3    // Function to count the operations required to reduce either of the two numbers to zero
4    // by repeatedly subtracting the smaller one from the larger one.
5    int countOperations(int num1, int num2) {
6        int operationsCount = 0; // Initialize counter for operations
7
8        // Continue the loop until either num1 or num2 becomes zero
9        while (num1 != 0 && num2 != 0) {
10            // If num1 is greater than num2, then swap them so that num1 always has the smaller value
11            if (num1 > num2) {
12                std::swap(num1, num2);
13            }
14
15            // Subtract num1 from num2 (num2 is guaranteed to be the larger or equal number here)
16            num2 -= num1;
17
18            // Increment the operations counter since a valid subtraction operation was performed
19            ++operationsCount;
20        }
21
22        // Return the total number of operations performed
23        return operationsCount;
24    }
25};
26

1function countOperations(num1: number, num2: number): number {
2    let operationsCount = 0; // Initialize a counter to track the number of operations performed
3  
4    // Continue the process until either of the numbers becomes zero
5    while (num1 !== 0 && num2 !== 0) {
6        // Set num1 to the smaller of the two numbers
7        // Subtract the smaller number from the larger and set this as the new value of num2
8        [num1, num2] = [Math.min(num1, num2), Math.abs(num1 - num2)];
9      
10        operationsCount++; // Increment the counter after each operation
11    }
12  
13    return operationsCount; // Return the number of operations performed when the loop ends
14}
15

Time and Space Complexity

Time Complexity

The time complexity of the given algorithm is O(num1 + num2). This is because in each iteration of the while loop, the algorithm subtracts the smaller number from the larger one, guaranteeing that at least one of the numbers is reduced by a proportion of its value. In the worst case, if num1 and num2 are consecutive Fibonacci numbers (which is the worst-case scenario for this type of subtraction loop), it will take a number of steps equal to the smaller number.

However, the actual number of operations depends on the values of num1 and num2. If num1 is much smaller than num2, then num2 will be reduced very slowly, leading to a high number of operations approaching num2 / num1. Conversely, if num1 is comparable to num2, the number of operations decreases.

Space Complexity

The space complexity of the algorithm is O(1), since it uses a constant amount of space. The variables ans, num1, and num2 are the only variables that are being modified and stored during the execution, and their space requirement does not depend on the input size. The algorithm operates directly on these variables and does not allocate any additional space that scales with the input.

Learn more about how to find time and space complexity quickly using problem constraints.

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Question 1 out of 10

How does merge sort divide the problem into subproblems?

Divide the array into two equal halves by index

Divide the array into two based on whether an element is smaller than an arbitrary value

Divide the array into a stray element and the rest of the array

Merge sort does not use divide and conquer

I don’t know