Assign other value to a variable from two possible values
Last Updated :
06 Mar, 2023
Suppose a variable x can have only two possible values a and b, and you wish to assign to x the value other than its current one. Do it efficiently without using any conditional operator.
Note: We are not allowed to check current value of x.
Examples:
Input : a = 10, b = 15, x = a
Output : x = 15
Explanation x = 10, currently x has value of a (which is 10), we need to change it to 15.
Input : a = 9, b = 11, x = b
Output : x = 9
We could solve this problem using if condition, but we are not allowed to do that.
if (x == a)
x = b;
else x = a;
We could have used Ternary operator, it also checks the current value of x and based on that it assigns new value. So we cannot use this approach as well
x = x == a ? b : a;
But, we are not allowed to check value of x, so none of the above solutions work.
Solution 1: Using arithmetic operators only we can perform this operation
x = a + b - x
This way the content of x will alternate between a and b every time it gets executed
C++
#include <bits/stdc++.h>
using namespace std;
void alternate(int& a, int& b, int& x)
{
x = a + b - x;
}
int main()
{
int a = -10;
int b = 15;
int x = a;
cout << "x is : " << x;
alternate(a, b, x);
cout << "\nAfter change ";
cout << "\nx is : " << x;
}
|
Java
import java.util.*;
class solution
{
static void alternate(int a, int b, int x)
{
x = a + b - x;
System.out.println("After change"+"\n"+" x is : "+x);
}
public static void main(String args[])
{
int a = -10;
int b = 15;
int x = a;
System.out.println("x is : "+x);
alternate(a, b, x);
}
}
|
Python3
def alternate(a,b,x):
x = a+b-x
print("After change x is:",x)
if __name__=='__main__':
a = -10
b = 15
x = a
print("x is:",x)
alternate(a,b,x)
|
C#
using System;
class gfg
{
public void alternate(ref int a, ref int b, ref int x)
{
x = a + b - x;
}
}
class geek
{
public static int Main()
{
gfg g = new gfg();
int a = -10;
int b = 15;
int x = a;
Console.WriteLine("x is : {0}" , x);
g.alternate(ref a, ref b, ref x);
Console.WriteLine ("After change ");
Console.WriteLine("x is : {0}", x);
return 0;
}
}
|
PHP
<?php
function alternate (&$a, &$b, &$x)
{
$x = $a + $b - $x;
}
$a = -10;
$b = 15;
$x = $a;
echo "x is : ", $x;
alternate($a, $b, $x);
echo "\nAfter change ";
echo "\nx is : ", $x;
?>
|
Javascript
<script>
function alternate(a , b , x) {
x = a + b - x;
document.write("After change" + "<br/>" + " x is : " + x);
}
var a = -10;
var b = 15;
var x = a;
document.write("x is : " + x+"<br/>");
alternate(a, b, x);
</script>
|
Output:
x is : -10
After change
x is : 15
Time Complexity: The time complexity of this approach is O(1)
Space Complexity: The space complexity of this approach is O(1)
Solution 2: A better and efficient approach is using the bitwise XOR operation.
x = a^b^x
C++
#include <bits/stdc++.h>
using namespace std;
void alternate(int& a, int& b, int& x)
{
x = a ^ b ^ x;
}
int main()
{
int a = -10;
int b = 15;
int x = a;
cout << "x is : " << x;
alternate(a, b, x);
cout << "\nAfter exchange ";
cout << "\nx is : " << x;
return 0;
}
|
Java
class GFG {
static int alternate(int a, int b, int x) {
return x = a ^ b ^ x;
}
public static void main(String[] args) {
int a = -10;
int b = 15;
int x = a;
System.out.print("x is : " + x);
x = alternate(a, b, x);
System.out.print("\nAfter exchange ");
System.out.print("\nx is : " + x);
}
}
|
Python3
def alternate(a, b, x):
x = a ^ b ^ x
print("After exchange")
print("x is", x)
a = -10
b = 15
x = a
print("x is", x)
alternate(a, b, x)
|
C#
using System;
public class GFG {
static int alternate(int a, int b, int x) {
return x = a ^ b ^ x;
}
public static void Main() {
int a = -10;
int b = 15;
int x = a;
Console.Write("x is : " + x);
x = alternate(a, b, x);
Console.Write("\nAfter exchange ");
Console.Write("\nx is : " + x);
}
}
|
PHP
<?php
function alternate(&$a, &$b, &$x)
{
$x = $a ^ $b ^ $x;
}
$a = -10;
$b = 15;
$x = $a;
echo "x is : ", $x;
alternate($a, $b, $x);
echo "\nAfter exchange ";
echo "\nx is : ", $x;
?>
|
Javascript
<script>
function alternate(a , b , x) {
return x = a ^ b ^ x;
}
var a = -10;
var b = 15;
var x = a;
document.write("x is : " + x);
x = alternate(a, b, x);
document.write("<br/>After exchange ");
document.write("<br/>x is : " + x);
</script>
|
Output:
x is : -10
After exchange
x is : 15
Time Complexity: The time complexity of this approach is O(1)
Space Complexity: The space complexity of this approach is O(1)
Solution 3:
Using the multiplication operator, we can perform the operation:
x = a * b / x
This way the content of x will alternate between a and b.
This approach has only one step:
Step 1: x = a * b / x
C++
#include <bits/stdc++.h>
using namespace std;
void alternate(int& a, int& b, int& x)
{
x = a * b / x;
}
int main()
{
int a = -10;
int b = 15;
int x = a;
cout << "x is : " << x;
alternate(a, b, x);
cout << "\nAfter change ";
cout << "\nx is : " << x;
}
|
Java
import java.util.*;
public class Main {
public static int alternate(int a, int b, int x) {
x = a * b / x;
return x;
}
public static void main(String[] args) {
int a = -10;
int b = 15;
int x = a;
System.out.println("x is : " + x);
x=alternate(a, b, x);
System.out.println("\nAfter change ");
System.out.println("x is : " + x);
}
}
|
Python3
def alternate(a, b, x):
x = a * b // x
return x
def main():
a = -10
b = 15
x = a
print("x is :", x)
x = alternate(a, b, x)
print("\nAfter change ")
print("x is :", x)
if __name__ == '__main__':
main()
|
C#
using System;
public class Program
{
static void alternate(ref int a, ref int b, ref int x)
{
x = a * b / x;
}
public static void Main()
{
int a = -10;
int b = 15;
int x = a;
Console.WriteLine("x is : " + x);
alternate(ref a, ref b, ref x);
Console.WriteLine("\nAfter change ");
Console.WriteLine("x is : " + x);
}
}
|
Javascript
function alternate(a, b,x )
{
x = a * b / x;
return x;
}
let a = -10;
let b = 15;
let x = a;
console.log("x is : " + x);
x = alternate(a, b, x);
console.log("After change ");
console.log("x is : " + x);
|
Output:
x is : -10
After exchange
x is : 15
Time Complexity: The time complexity of this approach is O(1)
Auxiliary Space: The space complexity of this approach is O(1)
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