Hello Everyone,
Given an array A of n elements, sort the array according to 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:
- C++
// C++ program to rearrange the elements
// in array such that even positioned are
// greater than odd positioned elements
#include <bits/stdc++.h>
using namespace std;
void assign( int a[], int n)
{
// Sort the array
sort(a, a + n);
int ans[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++)
cout << ans[i] << " " ;
}
// Driver Code
int main()
{
int A[] = { 1, 3, 2, 2, 5 };
int n = sizeof (A) / sizeof (A[0]);
assign(A, n);
return 0;
}
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:
- C++
// C++ program to rearrange the elements
// in the array such that even positioned are
// greater than odd positioned elements
#include <bits/stdc++.h>
using namespace std;
// swap two elements
void swap( int * a, int * b)
{
int temp = *a;
*a = *b;
*b = temp;
}
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(&arr[i - 1], &arr[i]);
}
// if index is odd
else {
if (arr[i] < arr[i - 1])
swap(&arr[i - 1], &arr[i]);
}
}
}
int main()
{
int n = 5;
int arr[] = { 1, 3, 2, 2, 5 };
rearrange(arr, n);
for ( int i = 0; i < n; i++)
cout << arr[i] << " " ;
cout << "\n" ;
return 0;
}
Output:
1 3 2 5 2
Time Complexity: O(n)
Auxiliary Space: O(1)