Rearrange an array in maximum minimum form (O(1) extra space)

Software Development
#1

Hello Everyone,

In this post a solution that requires O(n) time and O(1) extra space is discussed. The idea is to use multiplication and modular trick to store two elements at an index.

even index : remaining maximum element. odd index : remaining minimum element. max_index : Index of remaining maximum element (Moves from right to left) min_index : Index of remaining minimum element (Moves from left to right) Initialize: max_index = ‘n-1’ min_index = 0 max_element = arr[max_index] + 1 //can be any element which is more than the maximum value in array For i = 0 to n-1 If ‘i’ is even arr[i] += arr[max_index] % max_element * max_element max_index-- ELSE // if ‘i’ is odd arr[i] += arr[min_index] % max_element * max_element min_index++

How does expression “arr[i] += arr[max_index] % max_element * max_element” work ?
The purpose of this expression is to store two elements at index arr[i]. arr[max_index] is stored as multiplier and “arr[i]” is stored as remainder. For example in {1 2 3 4 5 6 7 8 9}, max_element is 10 and we store 91 at index 0. With 91, we can get original element as 91%10 and new element as 91/10.

Below implementation of above idea:

// C++ program to rearrange an array in minimum

// maximum form

#include <bits/stdc++.h>

using namespace std;

// Prints max at first position, min at second position

// second max at third position, second min at fourth

// position and so on.

void rearrange( int arr[], int n)

{

// initialize index of first minimum and first

// maximum element

int max_idx = n - 1, min_idx = 0;

// store maximum element of array

int max_elem = arr[n - 1] + 1;

// traverse array elements

for ( int i = 0; i < n; i++) {

// at even index : we have to put maximum element

if (i % 2 == 0) {

arr[i] += (arr[max_idx] % max_elem) * max_elem;

max_idx--;

}

// at odd index : we have to put minimum element

else {

arr[i] += (arr[min_idx] % max_elem) * max_elem;

min_idx++;

}

}

// array elements back to it's original form

for ( int i = 0; i < n; i++)

arr[i] = arr[i] / max_elem;

}

// Driver program to test above function

int main()

{

int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };

int n = sizeof (arr) / sizeof (arr[0]);

cout << "Original Arrayn" ;

for ( int i = 0; i < n; i++)

cout << arr[i] << " " ;

rearrange(arr, n);

cout << "\nModified Array\n" ;

for ( int i = 0; i < n; i++)

cout << arr[i] << " " ;

return 0;

}

Output :

Original Array 1 2 3 4 5 6 7 8 9 Modified Array 9 1 8 2 7 3 6 4 5