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++
// CPP program to change value of x// according to its current value.#include <bits/stdc++.h>using namespace std;// Function to alternate the valuesvoid alternate(int& a, int& b, int& x){ x = a + b - x;}// Main functionint 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
// Java program to change value of x// according to its current value.import java.util.*;class solution{// Function to alternate the valuesstatic void alternate(int a, int b, int x){ x = a + b - x; System.out.println("After change"+"\n"+" x is : "+x);}// Main functionpublic 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
# Python3 program to change value # of x according to its current value. # Function to alternate the values def alternate(a,b,x): x = a+b-x print("After change x is:",x)# Driver codeif __name__=='__main__': a = -10 b = 15 x = a print("x is:",x) alternate(a,b,x)# This code is contributed by # Shrikant13 |
C#
// C# program to change value of x// according to its current value.using System;class gfg{ // Function to alternate the values public void alternate(ref int a, ref int b, ref int x) //'ref' indicates the references { x = a + b - x; }}// Main functionclass 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; }}//This code is contributed by Soumik |
PHP
<?php// PHP program to change value of x // according to its current value. // Function to alternate the values function alternate (&$a, &$b, &$x) { $x = $a + $b - $x; } // Driver Code$a = -10; $b = 15; $x = $a; echo "x is : ", $x; alternate($a, $b, $x); echo "\nAfter change "; echo "\nx is : ", $x; // This code is contributed by ajit.?> |
Javascript
<script>// javascript program to change value of x// according to its current value. // Function to alternate the values function alternate(a , b , x) { x = a + b - x; document.write("After change" + "<br/>" + " x is : " + x); } // Main function var a = -10; var b = 15; var x = a; document.write("x is : " + x+"<br/>"); alternate(a, b, x);// This code is contributed by todaysgaurav</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++
// CPP program to change value of x// according to its current value.#include <bits/stdc++.h>using namespace std;// Function to alternate the valuesvoid alternate(int& a, int& b, int& x){ x = a ^ b ^ x;}// Main functionint 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
// Java program to change value of x// according to its current value.class GFG {// Function to alternate the values static int alternate(int a, int b, int x) { return x = a ^ b ^ x; }// Main function 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); }} // This code is contributed by 29AjayKumar |
Python3
# Python3 program to change value of x# according to its current value.# Function to alternate the valuesdef alternate(a, b, x): x = a ^ b ^ x print("After exchange") print("x is", x)# Driver codea = -10b = 15x = aprint("x is", x)alternate(a, b, x)# This code is contributed # by Shrikant13 |
C#
// C# program to change value of x // according to its current value. using System;public class GFG { // Function to alternate the values static int alternate(int a, int b, int x) { return x = a ^ b ^ x; } // Main function 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); } } /*This code is contributed by Rajput-Ji*/ |
PHP
<?php// PHP program to change value of x // according to its current value. // Function to alternate the values function alternate(&$a, &$b, &$x) { $x = $a ^ $b ^ $x; } // Driver Code$a = -10; $b = 15; $x = $a; echo "x is : ", $x; alternate($a, $b, $x); echo "\nAfter exchange "; echo "\nx is : ", $x; // This code is contributed // by akt_mit?> |
Javascript
<script>// Javascript program to change value of x// according to its current value.// Function to alternate the values function alternate(a , b , x) { return x = a ^ b ^ x; } // Main function 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);// This code contributed by Rajput-Ji </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++
// CPP program to change value of x// according to its current value.#include <bits/stdc++.h>using namespace std;// Function to alternate the valuesvoid alternate(int& a, int& b, int& x){ x = a * b / x;}// Main functionint 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;}//This code is contributed by phasing17 |
Java
import java.util.*;public class Main { // Function to alternate the values public static int alternate(int a, int b, int x) { x = a * b / x; return x; } // Main function 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
# Function to alternate the valuesdef alternate(a, b, x): x = a * b // x return x# Main functiondef 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#
// C# program to change value of x// according to its current value.using System;public class Program{ // Function to alternate the values static void alternate(ref int a, ref int b, ref int x) { x = a * b / x; } // Main function 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); }}// This code is contributed by bhardwajji |
Javascript
// JavaScript program to change value of x// according to its current value.// Function to alternate the valuesfunction alternate(a, b,x ){ x = a * b / x; return x; }// Main functionlet 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);// This code is contributed by phasing17 |
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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