Minimize the maximum absolute difference of adjacent elements in a circular array

Given a circular array arr of N integers, the task is to minimize the maximum absolute difference of adjacent elements of the array without any removals.

Examples:

Input: arr[] = {1, 3, 10, 2, 0, 9, 6}
Output: {0, 2, 6, 10, 9, 3, 1}
Explanation: In the above example, the maximum difference between adjacent elements is 6, which is between 9 and 3. Other orderings won’t be able to further minimize it.

Input: arr[] = {1, 2, 3, 4, 5, 6}
Output: {1, 3, 5, 6, 4, 2}
Example: The maximum difference is 2 between (1, 3) and (3, 5) and (6, 4) and (4, 2).

Approach
In order to solve the problem, just displaying the sorted array would lead to an incorrect solution as it is treated as a circular array. After sorting, the last and first indexed elements are the highest and lowest elements in the array respectively. Thus, the maximum difference between adjacent elements can be further minimized. So, after sorting, we need to reorder the sorted array such that the even indexed elements precede the odd indexed elements of the array and arrange the odd indexed elements in reverse order.

Illustration: For the given array arr[] = {1, 3, 10, 2, 0, 9, 6}, the sorted array will be {0, 1, 2, 3, 6, 9, 10}. The maximum difference between adjacent elements in the cirular array is |10 – 0| = 10. After reordering the array based on the above approach, we get the array to be {0, 2, 6, 10, 9, 3, 1}. Thus, the maximum difference is now minimized to |9 – 3| = 6

Below code is the implementation of the above approach:

 // C++ Program to minimize the // maximum absolute difference // between adjacent elements // of the circular array   #include using namespace std;   #define ll long long   // Function to print the reordered array // which minimizes thee maximum absolute // difference of adjacent elements void solve(vector& arr, int N) {     // Sort the given array     sort(arr.begin(), arr.end());     // Reorder the array     int fl = 1,k=0;     for(int i=0;i<=N/2;i++)     {         if((i%2 && fl) || !fl)         {             int x = arr[i];             arr.erase(arr.begin() + i);             arr.insert(arr.begin() + N - 1 - k, x);             k++;             fl = 0;         }     }     // Print the new ordering     for (int i : arr)         cout << i << " "; }     // Driver code int main() {     int N = 7;     vector arr = {1, 3, 10, 2, 0, 9, 6};     solve(arr, N);           return 0; }   // this code is contributed by divyanshu gupta

 // Java program to minimize the // maximum absolute difference // between adjacent elements // of the circular array import java.util.*;   class GFG{   // Function to print the reordered array // which minimizes thee maximum absolute // difference of adjacent elements static void solve(Vector arr, int N) {           // Sort the given array     Collections.sort(arr);           // Reorder the array     int fl = 1, k = 0;           for(int i = 0; i <= N / 2; i++)     {         if ((i % 2 != 0 && fl != 0) || fl == 0)         {             int x = arr.get(i);             arr.remove(i);             arr.add( N - 1 - k, x);             k++;             fl = 0;         }     }           // Print the new ordering     for(int i : arr)         System.out.print(i + " "); }   // Driver code public static void main(String[] args) {     int N = 7;     Vector arr = new Vector<>();           arr.add(1);     arr.add(3);     arr.add(10);     arr.add(2);     arr.add(0);     arr.add(9);     arr.add(6);           solve(arr, N); } }   // This code is contributed by Amit Katiyar

 # Python3 Program to minimize the # maximum absolute difference # between adjacent elements # of the circular array   # Function to print the reordered array # which minimizes thee maximum absolute # difference of adjacent elements def solve(arr, N):           # Sort the given array     arr.sort(reverse = False)           # Reorder the array     fl = 1     k=0     for i in range(N // 2 + 1):         if((i % 2 and fl) or fl == 0):             x = arr[i]             arr.remove(arr[i])             arr.insert(N - 1 - k, x)             k += 1             fl = 0                   # Print the new ordering     for i in arr:         print(i, end = " ")   # Driver code if __name__ == '__main__':           N = 7           arr = [ 1, 3, 10, 2, 0, 9, 6 ]     solve(arr, N)     # This code is contributed by Samarth

Output:
0 2 6 10 9 3 1

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