Given an array A of n elements, sort the array according to the following relations :
A[i] >= A[i-1]
, if i is even.
A[i] <= A[i-1]
, if i is odd.
Print the resultant array.
Examples :
Input : A[] = {1, 2, 2, 1}
Output : 1 2 1 2
Explanation :
For 1st element, 1 1, i = 2 is even.
3rd element, 1 1, i = 4 is even.
Input : A[] = {1, 3, 2}
Output : 1 3 2
Explanation :
Here, the array is also sorted as per the conditions.
1 1 and 2 < 3.
Method 1 –
Observe that array consists of [n/2] even positioned elements. If we assign the largest [n/2] elements to the even positions and the rest of the elements to the odd positions, our problem is solved. Because element at the odd position will always be less than the element at the even position as it is the maximum element and vice versa. Sort the array and assign the first [n/2] elements at even positions.
Below is the implementation of the above approach:
// Java program to rearrange the elements
// in array such that even positioned are
// greater than odd positioned elements
import java.io.*;
import java.util.*;
class GFG {
static void assign( int a[], int n)
{
// Sort the array
Arrays.sort(a);
int ans[] = new int [n];
int p = 0 , q = n - 1 ;
for ( int i = 0 ; i < n; i++) {
// Assign even indexes with maximum elements
if ((i + 1 ) % 2 == 0 )
ans[i] = a[q--];
// Assign odd indexes with remaining elements
else
ans[i] = a[p++];
}
// Print result
for ( int i = 0 ; i < n; i++)
System.out.print(ans[i] + " " );
}
// Driver code
public static void main(String args[])
{
int A[] = { 1 , 3 , 2 , 2 , 5 };
int n = A.length;
assign(A, n);
}
}
Output:
1 5 2 3 2
Time Complexity: O(n * log n)
Auxiliary Space: O(n)
Method 2 –
One other approach is to traverse the array from the second element and swap the element with the previous one if the condition is not satisfied. This is implemented as follows:
// Java program to rearrange the elements
// in the array such that even positioned are
// greater than odd positioned elements
class GFG
{
public static void rearrange( int [] arr, int n)
{
for ( int i = 1 ; i < n; i++)
{
// if index is even
if (i % 2 == 0 )
{
if (arr[i] > arr[i - 1 ])
{
// swap two elements
int temp = arr[i];
arr[i] = arr[i - 1 ];
arr[i - 1 ] = temp;
}
}
// if index is odd
else
{
if (arr[i] < arr[i - 1 ])
{
// swap two elements
int temp = arr[i];
arr[i] = arr[i - 1 ];
arr[i - 1 ] = temp;
}
}
}
for ( int i = 0 ; i < n; i++)
{
System.out.print(arr[i] + " " );
}
}
// Driver code
public static void main(String []args)
{
int n = 5 ;
int arr[] = { 1 , 3 , 2 , 2 , 5 };
rearrange(arr, n);
}
}
Output:
1 3 2 5 2
Time Complexity: O(n)
Auxiliary Space: O(1)