Program to swap two numbers

Swapping two numbers without using a temporary variable

To swap two numbers without using a temporary variable, we use the arithmetic operator.

Pseudo Code

Algorithm
___________________________________________________________________
procedure swapping():
___________________________________________________________________
1.   Declare a, b, temp
2.   a=a+b
3.   b=a-b
4.   a=a-b
5.   Display the value of a and b
end procedure
___________________________________________________________________

Implementation in C++

#include <iostream>
using namespace std;
int main() {
   int a, b, temp;
   cin>>a>>b;
   cout<<"before swapping"<<endl;
   cout<<"a:"<<a<<endl;
   cout<<"b:"<<b<<endl;
   
   //swapping
   a = a+b;
   b = a-b;
   a = a-b;
   
   //displaying
   cout<<"after swapping"<<endl;
   cout<<"a:"<<a<<endl;
   cout<<"b:"<<b<<endl;
   
   return 0;
}

You can also try this code with Online C++ Compiler

Run Code

Output:

Sample Input: 
5
6

Sample Output:
before swapping
a:5
b:6
after swapping
a:6
b:5

Complexity Analysis

Time Complexity: O(1), Since no extra operation, constant time is taken.

Space complexity: O(1), no extra variable used. 

Swapping two numbers using XOR

To swap two numbers using XOR, the following concept is kept in mind. The XOR operator always returns the second number if the result of two numbers subjected to XOR is again XOR-ed with the first original number, and returns the first number if the result of two numbers on passing through the XOR operator is again XOR-ed with the second original number.

Example: Say a=5 and b=7
a=5^7=2
b=2^7=5
a=2^5=7.

PseudoCode

___________________________________________________________________
procedure swapping():
___________________________________________________________________
1.   Declare a, b, temp
2.   a=a^b
3.   b=a^b
4.   a=a^b
5.   Display the value of a and b
end procedure
___________________________________________________________________

Implementation in C++

#include <iostream>
using namespace std;
int main() {
   int a, b;
   cin>>a>>b;
   cout<<"before swapping"<<endl;
   cout<<"a:"<<a<<endl;
   cout<<"b:"<<b<<endl;
   
   //swapping
   a = a ^ b;
   b = a ^ b;
   a = a ^ b;
   
   //displaying
   cout<<"after swapping"<<endl;
   cout<<"a:"<<a<<endl;
   cout<<"b:"<<b<<endl;
   return 0;
}

You can also try this code with Online C++ Compiler

Run Code

Output:

Sample Input: 
5
6


Sample Output:
before swapping
a:5
b:6
after swapping
a:6
b:5

Complexity Analysis

Time Complexity: O(1), Since no extra operation, constant time is taken.

Space complexity: O(1), no extra variable used.

Try and compile with online c++ compiler.

You can also read about - Strong number in c

Frequently Asked Questions

What is an XOR operator?

XOR operator is a logical operator returning false if input bits are the same and returns true if input bits are different.

What is space complexity?

Space complexity defines the amount of extra space used in an algorithm.
 

What is time complexity?

Time complexity measures the time taken by a computer to run an algorithm. It is of three types: best time complexity, average time complexity and worst time complexity.

Conclusion

This article taught us how to write a program to swap two numbers.

We discussed its implementation using a temporary variable and without using a temporary variable.

We hope you could take away critical techniques like analysing problems by walking over the execution of the steps and finding out the pattern followed in most array problems.

Also read -  Decimal to Binary c++

Check out this problem - XOR Queries On Tree

Now, we recommend you practice problem sets based on arrays to master your fundamentals. You can get a wide range of questions similar to this on Coding Ninjas Studio. 

It's not the end. Must solve the problem of similar types. 

Happy Coding.