Check if it is possible to make x and y zero at same time with given operation

Given two numbers X and Y. The task is to check whether X and Y can be reduced to zero at the same time by performing the following operation any number of times: 

• Choose any natural number (say z) and reduce X and Y as one of the following at each operation: 
  1. X = X – z and Y = Y – 2*z
  2. X = X – 2*z and Y = Y – z

Example:

Input: X = 6, Y = 9 
Output: YES 
Explanation:
We can perform operation in following way: 
if z = 1, then 
X = X – 2*z = 6 – 2*(1) 
Y = Y – z = 9 – 1 
=> X = 4 & Y = 8
Now again if z = 4, then 
X = X – z = 4 – 4 
Y = Y – 2*z = 8 – 2*(4) 
=> X = 0 & Y = 0 
Therefore, X & Y become zero in 2 steps assuming z as 1 and 4 respectively. 

Input: X = 1, Y = 1 
Output: NO 
Explanation:
We don’t have any possible value for z such that X & Y can become zero simultaneously. 

Approach:
Below are the observation for the given problem statement:  

1. Since X and Y are updated to (X – z and Y – 2*z) or (X – 2*z and Y – z), therefore after n number of operations (X + Y) is updated to (X + Y – 3*n*z). Hence X and Y can be reduced to zero at simultaneously if (X+Y)%3 equals 0.
2. At each step one of the X or Y is reduced by 2*z. To reduced X and Y simultaneously zero it must satisfy this condition: max(X, Y)≤ 2*min(X, Y). 
   For Example: 

Let X = 6 and Y = 15 
Since (X+Y)%3 = (21%3) = 0 
As our first condition is satisfied, 
But by taking z = 6 
X = X – z = 6 – 6 = 0 
Y = Y – 2*z = 15 – 12 = 3 
Since Y is not less than or equals to 2*X, therefore X and Y cannot be reduced to zero at same time. 

If the above two conditions satisfy the values of X and Y, then X and Y can be reduced to 0 simultaneously.

Below is the implementation of the above approach: 

C++

// C++ program of the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to check if it is possible to
// make x and y can become 0 at same time
void canBeReduced(int x, int y)
{
    int maxi = max(x, y);
    int mini = min(x, y);

    // Check the given conditions
    if (((x + y) % 3) == 0 && maxi <= 2*mini)
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
}

// Driver Code
int main()
{
    int x = 6, y = 9;

    // Function Call
    canBeReduced(x, y);
    return 0;
}

Java

// Java program of the above approach
import java.util.*;

class GFG{

// Function to check if it is possible to
// make x and y can become 0 at same time
static void canBeReduced(int x, int y)
{
    int maxi = Math.max(x, y);
    int mini = Math.min(x, y);

    // Check the given conditions
    if (((x + y) % 3) == 0 && maxi <= 2*mini)
        System.out.print("YES" +"\n");
    else
        System.out.print("NO" +"\n");
}

// Driver Code
public static void main(String[] args)
{
    int x = 6, y = 9;

    // Function Call
    canBeReduced(x, y);
}
}

// This code is contributed by Rajput-Ji

C#

// C# program of the above approach
using System;

class GFG
{
    // Function to check if it is possible to
    // make x and y can become 0 at same time
    static void canBeReduced(int x, int y)
    {
        int maxi = Math.Max(x, y);
        int mini = Math.Min(x, y);

        // Check the given conditions
        if (((x + y) % 3) == 0 && maxi <= 2*mini)
            Console.WriteLine("YES");
        else
            Console.WriteLine("NO");
    }

    // Driver Code
    static void Main()
    {
        int x = 6, y = 9;

        // Function Call
        canBeReduced(x, y);
    }
}

// This code is contributed by shubhamsingh10

Python3

# Python 3 program of the above approach

# Function to check if it is possible to
# make x and y can become 0 at same time
def canBeReduced(x,y):
    maxi = max(x, y)
    mini = min(x, y)

    # Check the given conditions
    if (((x + y) % 3) == 0 and maxi <= 2*mini):
        print("YES")
    else:
        print("NO")

# Driver Code
if __name__ == '__main__':
    x = 6
    y = 9

    # Function Call
    canBeReduced(x, y)

# This code is contributed by Surendra_Gangwar

Javascript

<script>

// javascript program of the above approach

// Function to check if it is possible to
// make x and y can become 0 at same time
function canBeReduced(x , y)
{
    var maxi = Math.max(x, y);
    var mini = Math.min(x, y);

    // Check the given conditions
    if (((x + y) % 3) == 0 && maxi <= 2*mini)
        document.write("YES" +"\n");
    else
        document.write("NO" +"\n");
}

// Driver Code

var x = 6, y = 9;

// Function Call
canBeReduced(x, y);

// This code contributed by shikhasingrajput

</script>

Output:

YES

Time Complexity: O(1)