Check if X and Y can be made zero by using given operation any number of times
Last Updated :
21 Dec, 2021
Given two integers X and Y, the task is to check if these two integers can be made equal to 0 by using the given operation any number of times. An operation is described as follows –
- Choose an arbitrary integer Z.
- Update the values with either of the following values:
- X = X – 2*Z and Y = Y – 3*Z
- X = X – 3*Z and Y = Y – 2*Z
Examples:
Input: X = 6, Y = 9
Output: Yes
Explanation:
Operation 1: Choose Z = 3, X = 6 – 2*3 = 0 and Y = 9 – 3*3 = 0
Since X and Y can be made equal to 0 using 1 operation, the required answer is Yes.
Input: X = 33, Y = 27
Output: Yes
Explanation:
Operation 1: Choose Z = 9, X := 33 – 3*9 = 6 and Y := 27 – 2*9 = 9
Operation 2: Choose Z = 3, X := 6 – 2*3 = 0 and Y := 9 – 3*3 = 0
Since X and Y can be made equal to 0 using 2 operation, the required answer is Yes.
Approach: Let’s assume X ? Y. Then the answer is Yes if two following conditions holds:
- (X + Y) mod 5 = 0: because after each operation the value (X + Y) mod 5 does not change.
Let’s assume some arbitrary number Z has been chosen.
Therefore, the value (X + Y) will be changed to
((X - 3Z) + (Y - 2Z))
(X + Y - 5Z)
- For this value to be equal to 0, X + Y = 5Z. Therefore, on taking mod on both the sides, (X + Y) mod 5 has to be equal to 0.
- 3*X >= 2*Y so that the subtraction doesnt make the values of X and Y negative.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void ifPossible( int X, int Y)
{
if (X > Y)
swap(X, Y);
if ((X + Y) % 5 == 0 and 3 * X >= 2 * Y)
cout << "Yes" ;
else
cout << "No" ;
}
int main()
{
int X = 33, Y = 27;
ifPossible(X, Y);
return 0;
}
|
Java
class GFG
{
static void ifPossible( int X, int Y)
{
if (X > Y)
swap(X, Y);
if ((X + Y) % 5 == 0 && 3 * X >= 2 * Y)
System.out.print( "Yes" );
else
System.out.print( "No" );
}
static void swap( int x, int y)
{
int temp = x;
x = y;
y = temp;
}
public static void main(String[] args)
{
int X = 33 , Y = 27 ;
ifPossible(X, Y);
}
}
|
Python3
def ifPossible(X, Y):
if (X > Y):
X, Y = Y, X
if ((X + Y) % 5 = = 0 and 3 * X > = 2 * Y):
print ( "Yes" )
else :
print ( "No" )
X = 33
Y = 27
ifPossible(X, Y)
|
C#
using System;
class GFG
{
static void ifPossible( int X, int Y)
{
if (X > Y)
swap(X, Y);
if ((X + Y) % 5 == 0 && 3 * X >= 2 * Y)
Console.WriteLine( "Yes" );
else
Console.WriteLine( "No" );
}
static void swap( int x, int y)
{
int temp = x;
x = y;
y = temp;
}
public static void Main()
{
int X = 33, Y = 27;
ifPossible(X, Y);
}
}
|
Javascript
<script>
function ifPossible(X, Y)
{
if (X > Y)
{
var temp = X;
X = Y;
Y =temp;
}
if ((X + Y) % 5 == 0 && 3 * X >= 2 * Y)
document.write( "Yes" );
else
document.write( "No" );
}
var X = 33, Y = 27;
ifPossible(X, Y);
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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