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 circular 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 is the implementation of the above approach:
C++
// C++ Program to minimize the // maximum absolute difference // between adjacent elements // of the circular array#include <bits/stdc++.h>using namespace std;#define ll long long// Function to print the reordered array// which minimizes the maximum absolute// difference of adjacent elementsvoid solve(vector<int>& 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 codeint main(){ int N = 7; vector<int> arr = {1, 3, 10, 2, 0, 9, 6}; solve(arr, N); return 0;}// this code is contributed by divyanshu gupta |
Java
// Java program to minimize the // maximum absolute difference // between adjacent elements // of the circular arrayimport java.util.*;class GFG{// Function to print the reordered array// which minimizes the maximum absolute// difference of adjacent elementsstatic void solve(Vector<Integer> 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 codepublic static void main(String[] args){ int N = 7; Vector<Integer> 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
# Python3 Program to minimize the # maximum absolute difference # between adjacent elements # of the circular array# Function to print the reordered array# which minimizes the maximum absolute# difference of adjacent elementsdef 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 codeif __name__ == '__main__': N = 7 arr = [ 1, 3, 10, 2, 0, 9, 6 ] solve(arr, N)# This code is contributed by Samarth |
C#
// C# program to minimize the // maximum absolute difference // between adjacent elements // of the circular arrayusing System;using System.Collections.Generic;class GFG{// Function to print the // reordered array which // minimizes the maximum // absolute difference of // adjacent elementsstatic void solve(List<int> arr, int N){ // Sort the given array arr.Sort(); // 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[i]; arr.RemoveAt(i); arr.Insert(N - 1 - k, x); k++; fl = 0; } } // Print the new ordering foreach(int i in arr) Console.Write(i + " ");}// Driver codepublic static void Main(String[] args){ int N = 7; List<int> arr = new List<int>(); 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 Rajput-Ji |
Javascript
<script>// JavaScript Program to minimize the// maximum absolute difference// between adjacent elements// of the circular array// Function to print the reordered array// which minimizes the maximum absolute// difference of adjacent elementsfunction solve(arr, N){ // Sort the given array arr.sort((a,b)=>a-b) // Reorder the array let fl = 1 let k=0 for(let i=0;i<(Math.log(N / 2) + 1);i++){ if((i % 2 && fl) || fl == 0){ let x = arr[i] arr = arr.filter((y)=>y != x) arr.splice(N - 1 - k,0,x) k += 1 fl = 0 } } // Print the new ordering for(let i of arr) document.write(i," ")}// Driver codelet N = 7 let arr = [ 1, 3, 10, 2, 0, 9, 6 ]solve(arr, N)// This code is contributed by shinjanpatra</script> |
0 2 6 10 9 3 1
Time complexity: O(N2) where N is the size of the given array
Auxiliary space: O(1)
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