Algorithm

Instead of using an extra variable, we can manipulate the values directly using arithmetic operations or bitwise XOR. For example, if we add the two numbers and then use subtraction, we can extract the original values in swapped order. Alternatively, using XOR works because applying XOR twice with the same value reverts back the original number. Both methods avoid using additional storage, though arithmetic operations may risk overflow while XOR is safer.

Take two numbers as input
Add the second number into the first number
Subtract the updated first number with the second number and assign it to the first number
Subtract the updated first number with the second number again and assign it to the second number
Print the swapped numbers

#include <stdio.h>
class Solution {
public:
    // Function to swap two numbers without using a temporary variable
    void swapNumbers(int *a, int *b) {
        // Add the two numbers and store result in first
        *a = *a + *b;
        
        // Subtract updated first with second to get the original first in second
        *b = *a - *b;
        
        // Subtract updated first with new second to get the original second in first
        *a = *a - *b;
    }
};

// Driver code
int main() {
    // Declare two numbers
    int num1, num2;
    
    // Take input from the user
    printf("Enter two numbers: ");
    scanf("%d %d", &num1, &num2);
    
    // Create an object of Solution class
    Solution obj;
    
    // Call the swap function
    obj.swapNumbers(&num1, &num2);
    
    // Print the swapped numbers
    printf("After swapping: num1 = %d, num2 = %d\n", num1, num2);
    
    return 0;
}

Complexity Analysis

Time Complexity: O(1) Swapping is achieved with a fixed number of arithmetic operations (three steps: addition and two subtractions) or bitwise operations (three XOR operations). The time taken does not grow with input size.

Space Complexity: O(1) No extra variables are introduced (beyond the given two inputs). Thus, space usage is constant.