Sort an array according to absolute difference with given value using Functors

Given an array of n distinct elements and a number x, arrange array elements according to the absolute difference with x, i. e., the element having a 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 : x = 7, arr[] = {10, 5, 3, 9, 2}
Output : arr[] = {5, 9, 10, 3, 2}
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}

Approach:
The above problem has been explained in the below post:
Sort an array according to absolute difference with given value
Sort an array according to absolute difference with a given value “using constant extra space”

The above-mentioned solutions have a time complexity of O(nlogn) with an auxiliary space of O(n) and O(n^2) with auxiliary space of O(1) respectively. In this post, a solution with time complexity of O(nlogn) using an auxiliary space of O(1) is proposed.

The solution is based on Functors. Compare the absolute value of the difference of the given number, k with the array value arr[i] and returns true if the value of the first object is less than the second object. Using stable_sort order of equivalent values is preserved.

Below is the implementation of the above approach:

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// C++ program to sort an array according
// absolute difference with x.
#include <bits/stdc++.h>
using namespace std;
  
// Functor class to perform comparison
class compare {
private:
    int num;
  
public:
    compare(int n)
    {
        num = n;
    }
  
    // Overloads () operator to perform
    // the desired comparison
    int operator()(int arr_num1, int arr_num2)
    {
        return abs(num - arr_num1) <
                          abs(num - arr_num2);
    }
};
  
// Function to sort an array according to
// absolute difference with x.
void rearrange(int arr[], int n, int k)
{
    // stable_sort sorts all the values in
    // non-decreasing order and preserves the
    // order of elements with equal values
    stable_sort(arr, arr + n, compare(k));
}
  
// 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 k = 7;
  
    rearrange(arr, n, k);
  
    printArray(arr, n);
  
    return 0;
}

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Output:

5 9 10 3 2

Time Complexity : O(n Log n)
Auxiliary Space : O(1)



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