Given an array arr[], the task is to rearrange the array elements by swapping adjacent elements such that no element remains at same position after swapping.
Examples:
Input:_ arr[] = { 1, 2, 3, 4, 5 }_
Output:_ 2 1 5 3 4_
Explanation:
Adjacent elements are swapped as follows:
(1, 2 -> 2, 1)
(3, 4, 5 -> 5, 3, 4)
Input:_ arr[] = {1, 2, 3, 4}_
Output:_ 2 1 4 3_
Explanation:
Adjacent elements are swapped as follows:
1, 2 -> 2, 1
3, 4 -> 4, 3
Approach: The key observation in the problem is that there can be two cases for the arrays to swap the array elements:
- If length of the array is even then we can easily Swap 2 Variables without using 3rd variable for every pair of contiguous elements.
- If length of array is odd then we can do the same as above but the the last 3 elements will not form a pair, So will we can easily Swap those 3 Variables without using 4th variable.
Below is the implementation of the above approach:
- C++
- C#
// C++ program for the above approach
#include <bits/stdc++.h>
**using** **namespace** std;
// Function to print the array
**void** print(``**int** a[], **int** n)
{
// Loop to iterate over the
// elements of array
**for** (``**int** i = 0; i < n; i++)
cout << a[i] << " "``;
}
// Function to swap two variables without
// using the third variable
**void** swapTwo(``**int**``& x, **int**``& y)
{
// Store XOR of all in x
x = x ^ y;
// After this, y has value of x
y = x ^ y;
// After this, x has value of y
x = x ^ y;
}
// Function to swap three variables
// without using fourth variable
**void** swapThree(``**int**``& a, **int**``& b, **int**``& c)
{
// Store XOR of all in a
a = a ^ b ^ c;
// After this, b has value of a
b = a ^ b ^ c;
// After this, c has value of b
c = a ^ b ^ c;
// After this, a has value of c
a = a ^ b ^ c;
}
// Function that swaps n integers
**void** rearrangeArray(``**int** a[], **int** n)
{
**if** (n % 2 == 0) {
**for** (``**int** i = 0; i < n - 1; i += 2) {
// Swap 2 variables without
// using 3rd variable
swapTwo(a[i], a[i + 1]);
}
}
**else** {
**for** (``**int** i = 0; i < n - 3; i += 2) {
// Swap 2 variables without
// using 3rd variable
swapTwo(a[i], a[i + 1]);
}
// The last 3 elements will not form
// pair if n is odd
// Hence, swapp 3 variables without
// using 4th variable
swapThree(a[n - 1], a[n - 2], a[n - 3]);
}
// Print the array elements
print(a, n);
}
// Driver Code
**int** main()
{
// Given array arr[]
**int** arr[] = { 1, 2, 3, 4, 5 };
**int** n = **sizeof**``(arr) / **sizeof**``(arr[0]);
// Function Call
rearrangeArray(arr, n);
**return** 0;
}
Output:
2 1 4 5 3
Time Complexity:_ O(N)_
Auxiliary Space:_ O(1)_
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