# Lexicographically smallest permutation having maximum sum of differences between adjacent elements

• Difficulty Level : Medium
• Last Updated : 06 Jul, 2021

Given an array arr[] of size N, the task is to find the lexicographically smallest permutation of the given array such that the sum of difference between adjacent elements is maximum.

Examples:

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Input: arr[] = {1, 2, 3, 4, 5}
Output: 5 2 3 4 1
Explanation:
Sum of difference between adjacent elements = (5 – 2) + (2 – 3) + (3 – 4) + (4 – 1) = 4.
{5, 2, 3, 4, 1} is lexicographically the smallest array and 4 is the maximum sum possible.

Input: arr[] = {3, 4, 1}
Output: 4 3 1
Explanation:
Sum of the difference between adjacent elements = (4 – 3) + (3 – 1) = 3
{4, 3, 1} is the lexicographically smallest permutation of array elements possible. Maximum sum of adjacent elements possible is 3.

Naive Approach: The simplest approach is to generate all permutations of the given array and find the sum of each permutation while keeping track of the maximum sum obtained. In the end, print the lexicographically smallest permutation with the maximum sum.

Time Complexity: O(N * N!)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observation:

Let the required permutation of the array be {p1, p2, p3, …, pN – 2, pN – 1, pN}.
Sum of the difference of the adjacent elements, S = (p1-p2) + (p2-p3) +….+ (pN – 2 – pN – 1) + (pN – 1 – pN)
= p1 – pn

In order to maximize S, p1 should be the largest element and pN should be the smallest element in the given array arr[]. To build the lexicographically the smallest permutation, arrange the rest of the elements in increasing order. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find the lexicographically``// smallest permutation of an array such``// that the sum of the difference between``// adjacent elements is maximum``void` `maximumSumPermutation(vector<``int``>& arr)``{` `    ``// Stores the size of the array``    ``int` `N = arr.size();` `    ``// Sort the given array in``    ``// increasing order``    ``sort(arr.begin(), arr.end());` `    ``// Swap the first and last array elements``    ``swap(arr, arr[N - 1]);` `    ``// Print the required permutation``    ``for` `(``int` `i : arr) {` `        ``cout << i << ``" "``;``    ``}``}` `// Driver Code``int` `main()``{``    ``vector<``int``> arr = { 1, 2, 3, 4, 5 };``    ``maximumSumPermutation(arr);` `    ``return` `0;``}`

## Java

 `// Java program for the above approach``import` `java.io.*;``import` `java.util.*;``class` `GFG``{` `    ``// Function to find the lexicographically``    ``// smallest permutation of an array such``    ``// that the sum of the difference between``    ``// adjacent elements is maximum``    ``static` `void` `maximumSumPermutation(``int``[] arr)``    ``{` `        ``// Stores the size of the array``        ``int` `N = arr.length;` `        ``// Sort the given array in``        ``// increasing order``        ``Arrays.sort(arr);` `        ``// Swap the first and last array elements``          ``int` `temp = arr[``0``];``          ``arr[``0``] = arr[N - ``1``];``        ``arr[N - ``1``] = temp;` `        ``// Print the required permutation``        ``for` `(``int` `i : arr) {` `            ``System.out.print(i + ``" "``);``        ``}``    ``}` `    ``// Driver Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``1``, ``2``, ``3``, ``4``, ``5` `};``        ``maximumSumPermutation(arr);``    ``}``}` `// This code is contributed by Dharanendra L V.`

## Python3

 `# Python program for the above approach` `# Function to find the lexicographically``# smallest permutation of an array such``# that the sum of the difference between``# adjacent elements is maximum``def` `maximumSumPermutation(arr):``  ` `    ``# Stores the size of the array``    ``N ``=` `len``(arr);` `    ``# Sort the given array in``    ``# increasing order``    ``arr.sort();` `    ``# Swap the first and last array elements``    ``temp ``=` `arr[``0``];``    ``arr[``0``] ``=` `arr[N ``-` `1``];``    ``arr[N ``-` `1``] ``=` `temp;` `    ``# Print the required permutation``    ``for` `i ``in` `arr:``        ``print``(i, end ``=` `" "``);` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ``arr ``=` `[``1``, ``2``, ``3``, ``4``, ``5``];``    ``maximumSumPermutation(arr);` `# This code is contributed by 29AjayKumar`

## C#

 `// C# program to implement``// the above approach``using` `System;``class` `GFG{` `  ``// Function to find the lexicographically``  ``// smallest permutation of an array such``  ``// that the sum of the difference between``  ``// adjacent elements is maximum``  ``static` `void` `maximumSumPermutation(``int``[] arr)``  ``{` `    ``// Stores the size of the array``    ``int` `N = arr.Length;` `    ``// Sort the given array in``    ``// increasing order``    ``Array.Sort(arr);` `    ``// Swap the first and last array elements``    ``int` `temp = arr;``    ``arr = arr[N - 1];``    ``arr[N - 1] = temp;` `    ``// Print the required permutation``    ``foreach` `(``int` `i ``in` `arr)``    ``{``      ``Console.Write(i + ``" "``);``    ``}``  ``}`  `  ``// Driver Code``  ``public` `static` `void` `Main(String[] args)``  ``{``    ``int``[] arr = { 1, 2, 3, 4, 5 };``    ``maximumSumPermutation(arr);``  ``}``}` `// This code is contributed by sanjoy_62.`

## Javascript

 ``
Output:
`5 2 3 4 1`

Time Complexity: O(N*log N)
Auxiliary Space: O(1)

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