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Rearrange given Array such that each elements is not equal to mean of adjacent elements

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  • Difficulty Level : Hard
  • Last Updated : 25 Aug, 2021
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Given an array arr consisting of N unique integers, the task is to rearrange the array such that element at index i of array should not be mean of adjacent elements (i.e., of index i-1 and i+1). Any possible rearrangement can be returned.

Example:

Input:  arr = [5, 4, 3, 2, 1]
Output:  [5, 3, 4, 2, 1]
Explanation: In the input array:
Mean(5, 3) = (5 + 3)/2 = 4,
Mean(4, 2) = (4+ 2 )/2 = 3,
Mean(3, 1) = (3 + 1)/2 = 2.
After rearranging the array as [5, 3, 4, 2, 1], now no element is the mean of adjacent elements: (5 + 4)/2 ≠ 3, (3 + 2)/2 ≠ 4, (4 + 1)/2 ≠ 2

Input: arr = [6, 9, 12, 25, 50 75]
Output:  [6, 12, 9, 25, 50, 75 ]

Approach: The main observation to solve this problem is that for 3 numbers a, b, and c to satisfy the condition that b shouldn’t be the mean of a and c, [a, b, c] mustn’t be sorted. Therefore, this problem can be solved by following steps:

  • Iterate over the array from 1 to (N-1) 
  • Check whether (arr[i – 1] + arr[i + 1]) / 2 == arr[i]) 
  • If the condition is satisfied swap the elements arr[i] and arr[i+1]

Below is the implementation of the above approach:

C++




// C++ code for above implementation
#include <bits/stdc++.h>
using namespace std;
 
// Function to rearrange the array
void Rearrange(int arr[], int N)
{
 
    // Iterating for array
    for (int i = 1; i < (N - 1); i++) {
 
        // Checking whether the element i
        // is mean of i-1 and i+1
        if ((arr[i - 1] + arr[i + 1]) / 2 == arr[i]) {
 
            // Rearrange by swapping arr[i] and arr[i+1]
            swap(arr[i], arr[i + 1]);
        }
    }
 
    // Printing the output array
    for (int i = 0; i < N; i++) {
        cout << arr[i] << " ";
    }
}
 
 
// Driver code
int main()
{
    int arr[] = { 6, 9, 12, 25, 50, 75 };
    int N = sizeof(arr) / sizeof(int);
 
    // calling the function
    Rearrange(arr, N);
    return 0;
}


Java




// Java program for the above approach
 
import java.io.*;
 
class GFG {
  // Function to rearrange the array
   
static void Rearrange(int arr[], int N)
{
   
    // Iterating for array
    for (int i = 1; i < (N - 1); i++) {
   
        // Checking whether the element i
        // is mean of i-1 and i+1
        if ((arr[i - 1] + arr[i + 1]) / 2 == arr[i]) {
   
            // Rearrange by swapping arr[i] and arr[i+1]
            
          int temp = arr[i];
        arr[i] = arr[i + 1];
        arr[i + 1] = temp;
        }
    }
   
    // Printing the output array
    for (int i = 0; i < N; i++) {
         System.out.print(arr[i] +" ");
    }
}
   
// Driver code
    public static void main (String[] args) {
      int arr[] = { 6, 9, 12, 25, 50, 75 };
    int N = arr.length;
   
    // calling the function
    Rearrange(arr, N);
    }
}
// This code is contributed by Potta Lokesh


C#

// C# program for the above approach

using System;

class GFG {
// Function to rearrange the array

static void Rearrange(int []arr, int N)
{

// Iterating for array
for (int i = 1; i < (N – 1); i++) {

// Checking whether the element i
// is mean of i-1 and i+1
if ((arr[i – 1] + arr[i + 1]) / 2 == arr[i]) {

// Rearrange by swapping arr[i] and arr[i+1]

int temp = arr[i];
arr[i] = arr[i + 1];
arr[i + 1] = temp;
}
}

// Printing the output array
for (int i = 0; i < N; i++) {
Console.Write(arr[i] +" ");
}
}

// Driver code
public static void Main (String[] args)
{
int []arr = { 6, 9, 12, 25, 50, 75 };
int N = arr.Length;

// calling the function
Rearrange(arr, N);
}
}
// This code is contributed by shivanisinghss2110

Python3




# Python3 program for the above approach
 
# Function to rearrange the array
def Rearrange(arr, N) :
 
    # Iterating for array
    for i in range(1, N - 1) :
 
        # Checking whether the element i
        # is mean of i-1 and i+1
        if ((arr[i - 1] + arr[i + 1]) // 2 == arr[i]) :
 
            # Rearrange by swapping arr[i] and arr[i+1]
            arr[i], arr[i + 1] = arr[i + 1], arr[i];
 
    # Printing the output array
    for i in range(N) :
        print(arr[i],end= " " )
   
 
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 6, 9, 12, 25, 50, 75 ];
    N = len(arr);
 
    # calling the function
    Rearrange(arr, N);
     
    # This code is contributed by AnkThon


Javascript




<script>
// JavaScript program for the above approach
// Function to rearrange the array
 
function Rearrange(arr, N)
{
 
    // Iterating for array
    for (var i = 1; i < (N - 1); i++) {
 
        // Checking whether the element i
        // is mean of i-1 and i+1
        if ((arr[i - 1] + arr[i + 1]) / 2 == arr[i]) {
 
            // Rearrange by swapping arr[i] and arr[i+1]
             
        var temp = arr[i];
        arr[i] = arr[i + 1];
        arr[i + 1] = temp;
        }
    }
 
    // Printing the output array
    for (var i = 0; i < N; i++) {
        document.write(arr[i] +" ");
    }
}
 
// Driver code
var arr = [ 6, 9, 12, 25, 50, 75 ];
var N = arr.length;
 
// calling the function
Rearrange(arr, N);
     
// This code is contributed by shivanisinghss2110
 
</script>


Output

6 12 9 25 75 50 

Time Complexity: O(N)
Auxiliary Space: O(1)


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