Sort an array according to absolute difference with a given value “using constant extra space”

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, using constant extra space.
Note : If two or more elements are at equal distance arrange them in same sequence as in the given array.

Examples:

Input  : arr[] = {10, 5, 3, 9, 2}
x = 7
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  : arr[] = {1, 2, 3, 4, 5}
x = 6
Output : arr[] = {5, 4, 3, 2, 1}

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

The above problem has already been explained in a previous post here. It takes O(n log n) time and O(n) extra space. The below solution though has a relatively bad time complexity i.e O(n^2) but it does the work without using any additional space or memory.

The solution is a based on Insertion Sort . For every i (1<= i < n) we compare the absolute value of the difference of arr[i] with the given number x (Let this be 'diff' ). We then compare this difference with the difference of abs(arr[j]-x) where 0<= j < i (Let this if abdiff). If diff is greater than abdiff, we shift the values in the array to accommodate arr[i] in it's correct position.

C++

 // C++ program to sort an array based on absolute // difference with a given value x. #include using namespace std;    void arrange(int arr[], int n, int x) {     // Below lines are similar to insertion sort     for (int i = 1; i < n; i++) {         int diff = abs(arr[i] - x);            // Insert arr[i] at correct place         int j = i - 1;         if (abs(arr[j] - x) > diff) {             int temp = arr[i];             while (abs(arr[j] - x) > diff && j >= 0) {                 arr[j + 1] = arr[j];                 j--;             }             arr[j + 1] = temp;         }     } }    // Function to print the array void print(int arr[], int n) {     for (int i = 0; i < n; i++)         cout << arr[i] << " "; }    // Main Function int main() {     int arr[] = { 10, 5, 3, 9, 2 };     int n = sizeof(arr) / sizeof(arr);     int x = 7;        arrange(arr, n, x);     print(arr, n);        return 0; }

Java

 // Java program to sort an array based on absolute // difference with a given value x. class GFG {    static void arrange(int arr[], int n, int x) {     // Below lines are similar to insertion sort     for (int i = 1; i < n; i++) {         int diff = Math.abs(arr[i] - x);            // Insert arr[i] at correct place         int j = i - 1;         if (Math.abs(arr[j] - x) > diff)         {             int temp = arr[i];             while (j >= 0 && Math.abs(arr[j] - x) > diff)              {                 arr[j + 1] = arr[j];                 j--;             }             arr[j + 1] = temp;         }     } }    // Function to print the array static void print(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;        arrange(arr, n, x);     print(arr, n);     } }    // This code is contributed by PrinciRaj1992

Python 3

 # Python 3 program to sort an array  # based on absolute difference with # a given value x. def arrange(arr, n, x):        # Below lines are similar to      # insertion sort     for i in range(1, n) :         diff = abs(arr[i] - x)            # Insert arr[i] at correct place         j = i - 1         if (abs(arr[j] - x) > diff) :             temp = arr[i]             while (abs(arr[j] - x) >                         diff and j >= 0) :                 arr[j + 1] = arr[j]                 j -= 1                            arr[j + 1] = temp    # Function to print the array def print_1(arr, n):        for i in range(n):         print(arr[i], end = " ")    # Driver Code if __name__ == "__main__":            arr = [ 10, 5, 3, 9, 2 ]     n = len(arr)     x = 7        arrange(arr, n, x)     print_1(arr, n)    # This code is contributed by ita_c

C#

 // C# program to sort an array based on absolute // difference with a given value x. using System;    class GFG  {        static void arrange(int []arr, int n, int x)     {                    // Below lines are similar to insertion sort         for (int i = 1; i < n; i++)         {             int diff = Math.Abs(arr[i] - x);                // Insert arr[i] at correct place             int j = i - 1;             if (Math.Abs(arr[j] - x) > diff)             {                 int temp = arr[i];                 while (j >= 0 && Math.Abs(arr[j] - x) > diff)                  {                     arr[j + 1] = arr[j];                     j--;                 }                 arr[j + 1] = temp;             }         }     }        // Function to print the array     static void print(int []arr, int n)     {         for (int i = 0; i < n; i++)             Console.Write(arr[i] + " ");     }        // Driver code     public static void Main()      {         int []arr = { 10, 5, 3, 9, 2 };         int n = arr.Length;         int x = 7;            arrange(arr, n, x);         print(arr, n);     } }    // This code is contributed by 29AjayKumar

PHP

 \$diff)         {             \$temp = \$arr[\$i];             while (abs(\$arr[\$j] - \$x) >                         \$diff && \$j >= 0)              {                 \$arr[\$j + 1] = \$arr[\$j];                 \$j--;             }             \$arr[\$j + 1] = \$temp;         }     }     return \$arr; }    // Function to print the array function print_arr(\$arr, \$n) {     for (\$i = 0; \$i < \$n; \$i++)         echo \$arr[\$i] . " "; }    // Driver Code \$arr = array(10, 5, 3, 9, 2); \$n = sizeof(\$arr); \$x = 7;    \$arr1 = arrange(\$arr, \$n, \$x); print_arr(\$arr1, \$n);    // This code is contributed  // by Akanksha Rai ?>

Output:

5 9 10 3 2

Time Complexity: O(n^2) where n is the size of the array.
Auxiliary Space: O(1)

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