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}

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++ program to sort an array based on absolute // difference with a given value x. #include <bits/stdc++.h> 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[0]); int x = 7; arrange(arr, n, x); print(arr, n); return 0; }

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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