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Rearrange array such that even positioned are greater than odd

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Given an array arr[] of N elements, sort the array according to the following relations:  

  • arr[i] >= arr[i – 1], if i is even, âˆ€ 1 <= i < N
  • arr[i] <= arr[i – 1], if i is odd, âˆ€ 1 <= i < N

Print the resultant array.

Examples:  

Input: N = 4, arr[] = {1, 2, 2, 1}
Output: 2 1 2 1
Explanation:

  • For i = 1, arr[1] <= arr[0]. So, 1 <= 2.
  • For i = 2, arr[2] >= arr[1]. So, 2 >= 1.
  • For i = 3, arr[3] <= arr[2]. So, 1 <= 2.

Input: arr[] = {1, 3, 2}
Output: 3 1 2
Explanation:

  • For i = 1, arr[1] <= arr[0]. So, 1 <= 3.
  • For i = 2, arr[2] >= arr[1]. So, 2 >= 1.

Approach: To solve the problem, follow the below idea:

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++




#include <bits/stdc++.h>
using namespace std;
 
// function to rearrange the elements in array such that
// even positioned are greater than odd positioned elements
void assign(int arr[], int N)
{
    // Sort the array
    sort(arr, arr + N);
 
    int ans[N];
    int ptr1 = 0, ptr2 = N - 1;
    for (int i = 0; i < N; i++) {
        // Assign even indexes with maximum elements
        if (i % 2 == 0)
            ans[i] = arr[ptr2--];
        // Assign odd indexes with remaining elements
        else
            ans[i] = arr[ptr1++];
    }
 
    // Print result
    for (int i = 0; i < N; i++)
        cout << ans[i] << " ";
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 2, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    assign(arr, N);
    return 0;
}


C




#include <stdio.h>
#include <stdlib.h>
 
// Compare function for qsort
int cmpfunc(const void* a, const void* b)
{
    return (*(int*)a - *(int*)b);
}
 
// function to rearrange the elements in array such that
// even positioned are greater than odd positioned elements
void assign(int arr[], int N)
{
    // Sort the array
    qsort(arr, N, sizeof(int), cmpfunc);
 
    int ans[N];
    int ptr1 = 0, ptr2 = N - 1;
    for (int i = 0; i < N; i++) {
        // Assign even indexes with maximum elements
        if (i % 2 == 0)
            ans[i] = arr[ptr2--];
        // Assign odd indexes with remaining elements
        else
            ans[i] = arr[ptr1++];
    }
 
    // Print result
    for (int i = 0; i < N; i++)
        printf("%d ", ans[i]);
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 2, 1 };
    int N = sizeof(arr) / sizeof(arr[0]);
    assign(arr, N);
    return 0;
}
 
// This code is contributed by Sania Kumari Gupta


Java




import java.io.*;
import java.util.*;
 
class GFG {
    // function to rearrange the elements
    // in array such that even positioned are
    // greater than odd positioned elements
    static void assign(int arr[], int N)
    {
        // Sort the array
        Arrays.sort(arr);
 
        int ans[] = new int[N];
        int ptr1 = 0, ptr2 = N - 1;
        for (int i = 0; i < N; i++) {
 
            // Assign even indexes with maximum elements
            if (i % 2 == 0)
                ans[i] = arr[ptr2--];
 
            // Assign odd indexes with remaining elements
            else
                ans[i] = arr[ptr1++];
        }
 
        // Print result
        for (int i = 0; i < N; i++)
            System.out.print(ans[i] + " ");
    }
 
    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 1, 2, 2, 1 };
        int N = arr.length;
        assign(arr, N);
    }
}
 
// This code is contributed by Nikita Tiwari.


C#




using System;
 
class GFG {
    // function to rearrange the elements
    // in array such that even positioned are
    // greater than odd positioned elements
    static void assign(int[] arr, int N)
    {
        // Sort the array
        Array.Sort(arr);
 
        int[] ans = new int[N];
        int ptr1 = 0, ptr2 = N - 1;
        for (int i = 0; i < N; i++) {
 
            // Assign even indexes with maximum elements
            if (i % 2 == 0)
                ans[i] = arr[ptr2--];
 
            // Assign odd indexes with remaining elements
            else
                ans[i] = arr[ptr1++];
        }
 
        // Print result
        for (int i = 0; i < N; i++)
            Console.Write(ans[i] + " ");
    }
 
    // Driver code
    public static void Main()
    {
        int[] arr = { 1, 2, 2, 1 };
        int N = arr.Length;
        assign(arr, N);
    }
}
 
// This code is contributed by vt_m.


Javascript




<script>
    // function to rearrange the elements
    // in array such that even positioned are
    // greater than odd positioned elements
    function assign(arr, N)
    {
  
        // Sort the array
        arr.sort();
  
        let ans = [];
        let ptr1 = 0, ptr2 = N - 1;
        for (let i = 0; i < N; i++) {
  
            // Assign even indexes with maximum elements
            if (i % 2 == 0)
                ans[i] = arr[ptr2--];
  
            // Assign odd indexes with remaining elements
            else
                ans[i] = arr[ptr1++];
        }
  
        // Print result
        for (let i = 0; i < N; i++)
            document.write(ans[i] + " ");
    }
 
// Driver code
 
        let arr = [ 1, 2, 2, 1 ];
        let N = arr.length;
        assign(arr, N);
                             
</script>


PHP




<?php
 
   
// function to rearrange the
// elements in array such that
// even positioned are greater
// than odd positioned elements
function assign($arr, $N)
{
     
    // Sort the array
    sort($arr);
 
    $ptr1 = 0; $ptr2 = $N - 1;
    for ($i = 0; $i < $N; $i++)
    {
         
        // Assign even indexes
        // with maximum elements
        if ($i % 2 == 0)
            $ans[$i] = $arr[$ptr2--];
 
        // Assign odd indexes
        // with remaining elements
        else
            $ans[$i] = $arr[$ptr1++];
    }
 
    // Print result
    for ($i = 0; $i < $N; $i++)
        echo($ans[$i] . " ");
}
 
// Driver Code
$arr = array( 1, 2, 2, 1 );
$N = sizeof($arr);
assign($arr, $N);
 
// This code is contributed by Ajit.
?>


Python3




# function  to rearrange the
# elements in array such that
# even positioned are greater
# than odd positioned elements
def assign(arr, N):
     
    # Sort the array
    arr.sort()
     
    ans = [0] * N
    ptr1 = 0
    ptr2 = N - 1
    for i in range(N):
         
        # Assign even indexes with
        # maximum elements
        if i % 2 == 0:
            ans[i] = arr[ptr2]
            ptr2 = ptr2 - 1
         
        # Assign odd indexes with
        # remaining elements
        else:
            ans[i] = arr[ptr1]
            ptr1 = ptr1 + 1
             
    # Print result
    for i in range(N):
        print(ans[i], end = " ")
 
# Driver Code
arr = [ 1, 2, 2, 1]
N = len(arr)
assign(arr, N)
 
# This code is contributed by "Sharad_Bhardwaj".


Output

2 1 2 1 

Time Complexity: O(N * log N), where N is the size of input array arr[].
Auxiliary Space: O(N)

Approach 2: Rearranging array by swapping elements

One other approach is to traverse the array from the first element till N – 1 and swap the element with the next one if the condition is not satisfied. This is implemented as follows: 

C++




#include <iostream>
using namespace std;
 
// function to rearrange the elements in array such that
// even positioned are greater than odd positioned elements
void rearrange(int arr[], int N)
{
    for (int i = 0; i < N; i += 2) {
          // Compare it with the previous element
        if (i > 0 && arr[i - 1] > arr[i])
            swap(arr[i - 1], arr[i]);
         
          // Compare it with the next element
        if (i < N - 1 && arr[i + 1] > arr[i])
            swap(arr[i + 1], arr[i]);
    }
}
 
int main()
{
      // Sample Input
    int N = 4;
    int arr[] = { 1, 2, 2, 1 };
       
    rearrange(arr, N);
     
      for (int i = 0; i < N; i++)
        cout << arr[i] << " ";
    cout << "\n";
    return 0;
}


Java




import java.io.*;
 
class GFG {
    // function to rearrange the elements in array such that
    // even positioned are greater than odd positioned elements
    static void rearrange(int[] arr, int N) {
        for (int i = 0; i < N; i += 2) {
            // Compare it with the previous element
            if (i > 0 && arr[i - 1] > arr[i]) {
                int temp = arr[i - 1];
                arr[i - 1] = arr[i];
                arr[i] = temp;
            }
 
            // Compare it with the next element
            if (i < N - 1 && arr[i + 1] > arr[i]) {
                int temp = arr[i + 1];
                arr[i + 1] = arr[i];
                arr[i] = temp;
            }
        }
    }
 
    public static void main(String[] args) {
        // Sample Input
        int N = 4;
        int[] arr = { 1, 2, 2, 1 };
 
        rearrange(arr, N);
 
        for (int i = 0; i < N; i++)
            System.out.print(arr[i] + " ");
        System.out.println();
    }
}
 
// This code is contributed by avanitrachhadiya2155


C#




// C# program to rearrange the elements
// in the array such that even positioned are
// greater than odd positioned elements
using System;
class GFG {
    // function to rearrange the elements in array such that
    // even positioned are greater than odd positioned
    // elements
    static void Rearrange(int[] arr, int N)
    {
        for (int i = 0; i < N; i += 2) {
            // Compare it with the previous element
            if (i > 0 && arr[i - 1] > arr[i]) {
                int temp = arr[i - 1];
                arr[i - 1] = arr[i];
                arr[i] = temp;
            }
 
            // Compare it with the next element
            if (i < N - 1 && arr[i + 1] > arr[i]) {
                int temp = arr[i + 1];
                arr[i + 1] = arr[i];
                arr[i] = temp;
            }
        }
    }
 
    static void Main()
    {
        // Sample Input
        int N = 4;
        int[] arr = { 1, 2, 2, 1 };
 
        Rearrange(arr, N);
 
        for (int i = 0; i < N; i++)
            Console.Write(arr[i] + " ");
        Console.WriteLine();
    }
}
 
// This code is contributed by shivanisinghss2110


Javascript




// function to rearrange the elements in array such that
// even positioned are greater than odd positioned elements
function rearrange(arr, N) {
    for (let i = 0; i < N; i += 2) {
        // Compare it with the previous element
        if (i > 0 && arr[i - 1] > arr[i]) {
            [arr[i - 1], arr[i]] = [arr[i], arr[i - 1]];
        }
 
        // Compare it with the next element
        if (i < N - 1 && arr[i + 1] > arr[i]) {
            [arr[i], arr[i + 1]] = [arr[i + 1], arr[i]];
        }
    }
}
 
// Sample Input
const N = 4;
const arr = [1, 2, 2, 1];
 
rearrange(arr, N);
 
console.log(arr.join(" "));


Python3




# function to rearrange the elements in array such that
# even positioned are greater than odd positioned elements
 
 
def rearrange(arr, N):
    for i in range(0, N, 2):
        # Compare it with the previous element
        if i > 0 and arr[i - 1] > arr[i]:
            arr[i - 1], arr[i] = arr[i], arr[i - 1]
 
        # Compare it with the next element
        if i < N - 1 and arr[i + 1] > arr[i]:
            arr[i], arr[i + 1] = arr[i + 1], arr[i]
 
 
# Sample Input
N = 4
arr = [1, 2, 2, 1]
 
rearrange(arr, N)
 
for i in range(N):
    print(arr[i], end=" ")
print()


Output

4 2 3 2 5 

Time Complexity: O(N), where N is the size of input array arr[].
Auxiliary Space: O(1)



Last Updated : 29 Feb, 2024
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