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Sort elements of an array in increasing order of absolute difference of adjacent elements

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Given an array arr[], the task is to arrange the array in such a way that the absolute difference between the adjacent elements is in increasing order.
 

Examples: 

Input: arr[] = {8, 1, 2, 3, 0} 
Output: 2 3 1 8 0 
Explanation: 
|2-3| = 1, |3-1| = 2, |1-8| = 7, |8-0| = 8 
The absolute difference between the adjacent elements is in increasing order. 

Input: -1 -2 3 4 5 
Output: 3 4 -1 5 -2 
Explanation: 
|3-4| = 1, |4-(-1)| = 5, |-1-5| = 6, |5-(-2)| = 7 

Approach: The idea is to sort the elements of the array and then take the middle element of the array and then repeat this step until the array is not empty. Below is an illustration of the approach with the help of an example: 

Given Array be - {8, 1, 2, 3, 0}

After Sorting the array - {0, 1, 2, 3, 8}

     Array          Middle Ele     Output Array
---------------    -------------  ---------------
{0, 1, 2, 3, 8}    5//2 = 2, 2          2
{0, 1, 3, 8}       4//2 = 2, 3        2, 3
{0, 1, 8}          3//2 = 1, 1       1, 2, 3
{0, 8}             2//2 = 1, 8      1, 2, 3, 8
{0}                1//2 = 0, 0     1, 2, 3, 8, 0

C++




// C++ implementation to sort
// the elements of the array in
// such a way that absolute 
// difference between the adjacent
// elements are in increasing order
#include<bits/stdc++.h>
using namespace std;
      
// Function to sort the elements
// of the array by difference
void sortDiff(vector<int> arr, 
              int n) 
{   
    // Sorting the array
    sort(arr.begin(), arr.end());
      
    // Array to store the
    // elements of the array
    vector<int> out;
      
    // Iterating over the length
    // of the array to include 
    // each middle element of array
    while(n > 0)
    {      
        // Appending the middle 
        // element of the array
        out.push_back(arr[n / 2]);
        arr.erase(arr.begin() + n / 2);
        n = n - 1;
    }
    for(auto i : out)
        cout << i << " ";
}
  
// Driver Code
int main()
{
  vector<int> a = {8, 1, 2, 3, 0};
  int n = 5;
  sortDiff(a, n);
}
  
// This code is contributed by Chitranayal


Java




// Java implementation to sort
// the elements of the array in
// such a way that absolute 
// difference between the adjacent
// elements are in increasing order
import java.util.*;
  
class GFG{
      
// Function to sort the elements
// of the array by difference
static void sortDiff(Vector<Integer> arr, int n) 
{
      
    // Sorting the array
    Collections.sort(arr);
      
    // Array to store the
    // elements of the array
    Vector<Integer> out = new Vector<Integer>();
      
    // Iterating over the length
    // of the array to include 
    // each middle element of array
    while(n > 0)
    {
          
        // Appending the middle 
        // element of the array
        out.add(arr.get(n / 2));
        arr.remove(n / 2);
        n = n - 1;
    }
    for(int i : out)
        System.out.print(i + " ");
}
  
// Driver Code
public static void main(String[] args)
{
    Integer []a = { 8, 1, 2, 3, 0 };
    Vector<Integer> arr = new Vector<Integer>(Arrays.asList(a));
    int n = 5;
    sortDiff(arr, n);
}
}
  
// This code is contributed by 29AjayKumar


Python3




# Python implementation to sort
# the elements of the array in
# such a way that absolute 
# difference between the adjacent
# elements are in increasing order
  
# Function to sort the elements
# of the array by difference
def sortDiff(arr, n):
      
    # Sorting the array
    arr.sort()
      
    # Array to store the
    # elements of the array
    out = []
      
    # Iterating over the length
    # of the array to include 
    # each middle element of array
    while n:
          
        # Appending the middle 
        # element of the array
        out.append(arr.pop(n//2))
        n=n-1
      
    print(*out)
    return out
  
# Driver Code
if __name__ == "__main__":
    arr = [8, 1, 2, 3, 0]
      
    n = 5
    sortDiff(arr, n)


C#




// C# implementation to sort
// the elements of the array in
// such a way that absolute 
// difference between the adjacent
// elements are in increasing order
using System;
using System.Collections.Generic;
class GFG{
      
// Function to sort the elements
// of the array by difference
static void sortDiff(List<int> arr, int n) 
{
      
    // Sorting the array
    arr.Sort();
      
    // Array to store the
    // elements of the array
    List<int> Out = new List<int>();
      
    // Iterating over the length
    // of the array to include 
    // each middle element of array
    while(n > 0)
    {
          
        // Appending the middle 
        // element of the array
        Out.Add(arr[n / 2]);
        arr.RemoveAt(n / 2);
        n = n - 1;
    }
    foreach(int i in Out)
        Console.Write(i + " ");
}
  
// Driver Code
public static void Main(String[] args)
{
    int []a = { 8, 1, 2, 3, 0 };
    List<int> arr = new List<int>(a);
    int n = 5;
    sortDiff(arr, n);
}
}
  
// This code is contributed by sapnasingh4991


Javascript




<script>
// Javascript implementation to sort
// the elements of the array in
// such a way that absolute
// difference between the adjacent
// elements are in increasing order
  
// Function to sort the elements
// of the array by difference
function sortDiff(arr,n)
{
    // Sorting the array
    arr.sort(function(a,b){return a-b;});
       
    // Array to store the
    // elements of the array
    let out = [];
       
    // Iterating over the length
    // of the array to include
    // each middle element of array
    while(n > 0)
    {
           
        // Appending the middle
        // element of the array
        out.push(arr[(Math.floor(n / 2))]);
        arr.splice(Math.floor(n / 2),1);
        n = n - 1;
    }
    for(let i=0;i< out.length;i++)
        document.write(out[i] + " ");
}
  
// Driver Code
let arr=[8, 1, 2, 3, 0 ];
let n = 5;
sortDiff(arr, n);
  
  
// This code is contributed by avanitrachhadiya2155
</script>


Output: 

2 3 1 8 0

 

Time Complexity: O(n2) as erase function takes O(n) time in the worst case.
Auxiliary Space: O(n)



Last Updated : 24 Mar, 2023
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