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 satify 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++

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// 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;
}

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Java

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// 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

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C#

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// 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

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Python3

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# 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

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Output:

YES

Time Complexity: O(1)

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