Sort an array according to absolute difference with given value

Given an array of n distinct elements and a number x, arrange array elements according to the absolute difference with x, i. e., element having minimum difference comes first and so on.
Note : If two or more elements are at equal distance arrange them in same sequence as in the given array.

Examples :

Input : arr[] : x = 7, arr[] = {10, 5, 3, 9, 2}
Output : arr[] = {5, 9, 10, 3, 2}
Explanation:
7 - 10 = 3(abs)
7 - 5 = 2
7 - 3 = 4 
7 - 9 = 2(abs)
7 - 2 = 5
So according to the difference with X, 
elements are arranged as 5, 9, 10, 3, 2.

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

Input : x = 5, arr[] = {2, 6, 8, 3}   
Output :  arr[] = {6, 3, 2, 8}

Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

The idea is to use a self balancing binary search tree. We traverse input array and for every element, we find its difference with x and store the difference as key and element as value in self balancing binary search tree. Finally we traverse the tree and print its inorder traversal which is required output.

C++ Implementation :
In C++, self-balancing-bianry-search-tree is implemented by set, map and multimap. We can’t use set here as we have key value pairs (not only keys). We also can’t directly use map also as a single key can belong to multiple values and map allows a single value for a key. So we use multimap which stores key value pairs and can have multiple values for a key.

Store the values in the multimap with the difference with X as key.
In multiimap, the values will be already in sorted order according to key i.e. difference with X because it implements self-balancing-bianry-search-tree internally.
Update all the values of array with the values of map so that array has the required output.

C++

// C++ program to sort an array according absolute 
// difference with x. 
#include<bits/stdc++.h> 
using namespace std; 
  
// Function to sort an array according absolute 
// difference with x. 
void rearrange(int arr[], int n, int x) 
{ 
    multimap<int, int> m; 
    multimap<int ,int >:: iterator it; 
    // Store values in a map with the difference 
    // with X as key 
    for (int i = 0 ; i < n; i++) 
        m.insert(make_pair(abs(x-arr[i]),arr[i])); 
  
    // Update the values of array 
    int i = 0; 
    for (it = m.begin(); it != m.end(); it++) 
        arr[i++] = (*it).second ; 
} 
  
// Function to print the array 
void printArray(int arr[] , int n) 
{ 
    for (int i = 0 ; i < n; i++) 
        cout << arr[i] << " "; 
} 
  
// Driver code 
int main() 
{ 
    int arr[] = {10, 5, 3, 9 ,2}; 
    int n = sizeof(arr)/sizeof(arr[0]); 
    int x = 7; 
    rearrange(arr, n, x); 
    printArray(arr, n); 
    return 0; 
} 

Java

// Java program to sort an array according absolute  
// difference with x.  
import java.io.*; 
import java.util.*; 
  
class GFG  
{ 
  
    // Function to sort an array according absolute  
    // difference with x. 
    static void rearrange(int[] arr, int n, int x) 
    { 
            TreeMap<Integer, ArrayList<Integer>> m = new TreeMap<>(); 
              
            // Store values in a map with the difference  
            // with X as key 
            for (int i = 0; i < n; i++) 
            { 
                int diff = Math.abs(x - arr[i]); 
                if (m.containsKey(diff))  
                { 
                    ArrayList<Integer> al = m.get(diff); 
                    al.add(arr[i]); 
                    m.put(diff, al); 
                } 
                else 
                { 
                        ArrayList<Integer> al = new ArrayList<>(); 
                        al.add(arr[i]); 
                        m.put(diff,al); 
                } 
            } 
  
            // Update the values of array 
            int index = 0; 
            for (Map.Entry entry : m.entrySet())  
            { 
                ArrayList<Integer> al = m.get(entry.getKey()); 
                Collections.sort(al); 
                for (int i = 0; i < al.size(); i++) 
                        arr[index++] = al.get(i);  
            } 
    } 
  
    // Function to print the array 
    static void printArray(int[] arr, int n) 
    { 
            for (int i = 0; i < n; i++) 
                System.out.print(arr[i] + " "); 
    } 
  
    // Driver code 
    public static void main(String args[]) 
    { 
            int[] arr = {10, 5, 3, 9 ,2}; 
            int n = arr.length; 
            int x = 7; 
            rearrange(arr, n, x);  
            printArray(arr, n);  
    } 
} 
  
// This code is contributed by rachana soma 

Output:

5 9 10 3 2

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

This article is contributed by Sahil Chhabra. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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