Rearrange array such that arr[i] >= arr[j] if i is even and arr[i]<=arr[j] if i is odd and j < i

• Difficulty Level : Medium
• Last Updated : 26 Aug, 2021

Given an array of n elements. Our task is to write a program to rearrange the array such that elements at even positions are greater than all elements before it and elements at odd positions are less than all elements before it.
Examples: 
 

Input : arr[] = {1, 2, 3, 4, 5, 6, 7}Output : 4 5 3 6 2 7 1Input : arr[] = {1, 2, 1, 4, 5, 6, 8, 8}Output : 4 5 2 6 1 8 1 8

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The idea to solve this problem is to first create an auxiliary copy of the original array and sort the copied array. Now total number of even position in array with n elements will be floor(n/2) and remaining is the number of odd positions. Now fill the odd and even positions in the original array using the sorted array in the below manner: 
 

• Total odd positions will be n – floor(n/2). Start from (n-floor(n/2))th position in the sorted array and copy the element to 1st position of sorted array. Start traversing the sorted array from this position towards left and keep filling the odd positions in the original array towards right.
• Start traversing the sorted array starting from (n-floor(n/2)+1)th position towards right and keep filling the original array starting from 2nd position. 
   

Below is the implementation of above idea: 
 

• C++
• Java
• Python3
• C#
• PHP
• Javascript

// C++ program to rearrange the array

// as per the given condition

#include <bits/stdc++.h>

using namespace std;

// function to rearrange the array

void rearrangeArr(int arr[], int n)

{

    // total even positions

    int evenPos = n / 2;

    // total odd positions

    int oddPos = n - evenPos;

    int tempArr[n];

    // copy original array in an

    // auxiliary array

    for (int i = 0; i < n; i++)

        tempArr[i] = arr[i];

    // sort the auxiliary array

    sort(tempArr, tempArr + n);

    int j = oddPos - 1;

    // fill up odd position in original

    // array

    for (int i = 0; i < n; i += 2) {

        arr[i] = tempArr[j];

        j--;

    }

    j = oddPos;

    // fill up even positions in original

    // array

    for (int i = 1; i < n; i += 2) {

        arr[i] = tempArr[j];

        j++;

    }

    // display array

    for (int i = 0; i < n; i++)

        cout << arr[i] << " ";

}

// Driver code

int main()

{

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

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

    rearrangeArr(arr, size);

    return 0;

}

Output: 
 

4 5 3 6 2 7 1

Time Complexity: O( n logn ) 
Auxiliary Space: O(n)

Another Approach-

We can traverse the array the by defining two variables p and q and assign values from last.

if even index is there then we will give it max value other wise min value.

p =0 and q= end;

p will go ahead and q will decrease.

• C++

#include <bits/stdc++.h>

using namespace std;

int main(){

    int n,i,j,p,q;

    int a[]= {1, 2, 1, 4, 5, 6, 8, 8};

    n=sizeof(a)/sizeof(a[0]);

    int b[n];

    for(i=0;i<n;i++)

        b[i]=a[i];

    sort(b,b+n);

    p=0;q=n-1;

    for(i=n-1;i>=0;i--){

            if(i%2!=0){

            a[i]=b[q];

            q--;

            }

            else{

                a[i]=b[p];

                p++;

            }

    }

    for(i=0;i<n;i++){

        cout<<a[i]<<" ";

    }

    return 0;

}

 This algorithm will take 1 for loop less than the previous one.