# Maximize sum of consecutive differences in a circular array

Given an array of **n** elements. Consider array as circular array i.e element after a_{n} is a_{1}. The task is to find maximum sum of the difference between consecutive elements with rearrangement of array element allowed i.e after rearrangement of element find |a_{1} – a_{2}| + |a_{2} – a_{3}| + …… + |a_{n – 1} – a_{n}| + |a_{n} – a_{1}|.

**Examples: **

Input : arr[] = { 4, 2, 1, 8 } Output : 18 Rearrange given array as : { 1, 8, 2, 4 } Sum of difference between consecutive element = |1 - 8| + |8 - 2| + |2 - 4| + |4 - 1| = 7 + 6 + 2 + 3 = 18. Input : arr[] = { 10, 12, 15 } Output : 10

The idea is to use Greedy Approach and try to bring elements having greater difference closer.

Consider the sorted permutation of the given array a_{1}, a_{1}, a_{2},…., a_{n – 1}, a_{n} such that a_{1} < a_{2} < a_{3}…. < a_{n – 1} < a_{n}.

Now, to obtain the answer having maximum sum of difference between consecutive element, arrange element in following manner:

a_{1}, a_{n}, a_{2}, a_{n-1},…., a_{n/2}, a_{(n/2) + 1}

We can observe that the arrangement produces the optimal answer, as all a_{1}, a_{2}, a_{3},….., a_{(n/2)-1}, a_{n/2} are subtracted twice while a_{(n/2)+1} , a_{(n/2)+2}, a_{(n/2)+3},….., a_{n – 1}, a_{n} are added twice.

Note – a_{(n/2)+1} This term is considered only for even n because for odd n, it is added once and subtracted once and hence cancels out.

## C++

`// C++ program to maximize the sum of difference` `// between consecutive elements in circular array` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Return the maximum Sum of difference between` `// consecutive elements.` `int` `maxSum(` `int` `arr[], ` `int` `n)` `{` ` ` `int` `sum = 0;` ` ` `// Sorting the array.` ` ` `sort(arr, arr + n);` ` ` `// Subtracting a1, a2, a3,....., a(n/2)-1, an/2` ` ` `// twice and adding a(n/2)+1, a(n/2)+2, a(n/2)+3,.` ` ` `// ...., an - 1, an twice.` ` ` `for` `(` `int` `i = 0; i < n/2; i++)` ` ` `{` ` ` `sum -= (2 * arr[i]);` ` ` `sum += (2 * arr[n - i - 1]);` ` ` `}` ` ` `return` `sum;` `}` `// Driver Program` `int` `main()` `{` ` ` `int` `arr[] = { 4, 2, 1, 8 };` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` `cout << maxSum(arr, n) << endl;` ` ` `return` `0;` `}` |

## Java

`// Java program to maximize the sum of difference` `// between consecutive elements in circular array` `import` `java.io.*;` `import` `java.util.Arrays;` ` ` `class` `MaxSum` `{` ` ` `// Return the maximum Sum of difference between` ` ` `// consecutive elements.` ` ` `static` `int` `maxSum(` `int` `arr[], ` `int` `n)` ` ` `{` ` ` `int` `sum = ` `0` `;` ` ` ` ` `// Sorting the array.` ` ` `Arrays.sort(arr);` ` ` ` ` `// Subtracting a1, a2, a3,....., a(n/2)-1,` ` ` `// an/2 twice and adding a(n/2)+1, a(n/2)+2,` ` ` `// a(n/2)+3,....., an - 1, an twice.` ` ` `for` `(` `int` `i = ` `0` `; i < n/` `2` `; i++)` ` ` `{` ` ` `sum -= (` `2` `* arr[i]);` ` ` `sum += (` `2` `* arr[n - i - ` `1` `]);` ` ` `}` ` ` ` ` `return` `sum;` ` ` `}` ` ` `// Driver Program` ` ` `public` `static` `void` `main (String[] args)` ` ` `{` ` ` `int` `arr[] = { ` `4` `, ` `2` `, ` `1` `, ` `8` `};` ` ` `int` `n = arr.length;` ` ` `System.out.println(maxSum(arr, n));` ` ` `}` `}` `/*This code is contributed by Prakriti Gupta*/` |

## Python3

`# Python3 program to maximize the sum of difference` `# between consecutive elements in circular array` `# Return the maximum Sum of difference` `# between consecutive elements` `def` `maxSum(arr, n):` ` ` `sum` `=` `0` ` ` `# Sorting the array` ` ` `arr.sort()` ` ` `# Subtracting a1, a2, a3,....., a(n/2)-1, an/2` ` ` `# twice and adding a(n/2)+1, a(n/2)+2, a(n/2)+3,.` ` ` `# ...., an - 1, an twice.` ` ` `for` `i ` `in` `range` `(` `0` `, ` `int` `(n ` `/` `2` `)) :` ` ` `sum` `-` `=` `(` `2` `*` `arr[i])` ` ` `sum` `+` `=` `(` `2` `*` `arr[n ` `-` `i ` `-` `1` `])` ` ` `return` `sum` `# Driver Program` `arr ` `=` `[` `4` `, ` `2` `, ` `1` `, ` `8` `]` `n ` `=` `len` `(arr)` `print` `(maxSum(arr, n))` `# This code is contributed by Shreyanshi Arun.` |

## C#

`// C# program to maximize the sum of difference` `// between consecutive elements in circular array` `using` `System;` `class` `MaxSum {` ` ` ` ` `// Return the maximum Sum of difference` ` ` `// between consecutive elements.` ` ` `static` `int` `maxSum(` `int` `[] arr, ` `int` `n)` ` ` `{` ` ` `int` `sum = 0;` ` ` `// Sorting the array.` ` ` `Array.Sort(arr);` ` ` `// Subtracting a1, a2, a3, ....., a(n/2)-1,` ` ` `// an/2 twice and adding a(n/2)+1, a(n/2)+2,` ` ` `// a(n/2)+3, ....., an - 1, an twice.` ` ` `for` `(` `int` `i = 0; i < n / 2; i++) {` ` ` `sum -= (2 * arr[i]);` ` ` `sum += (2 * arr[n - i - 1]);` ` ` `}` ` ` `return` `sum;` ` ` `}` ` ` `// Driver Program` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[] arr = { 4, 2, 1, 8 };` ` ` `int` `n = arr.Length;` ` ` `Console.WriteLine(maxSum(arr, n));` ` ` `}` `}` `//This Code is contributed by vt_m.` |

## Javascript

`<script>` `// JavaScript program for the above approach` ` ` `// Return the maximum Sum of difference between` ` ` `// consecutive elements.` ` ` `function` `maxSum(arr, n)` ` ` `{` ` ` `let sum = 0;` ` ` ` ` `// Sorting the array.` ` ` `arr.sort();` ` ` ` ` `// Subtracting a1, a2, a3,....., a(n/2)-1,` ` ` `// an/2 twice and adding a(n/2)+1, a(n/2)+2,` ` ` `// a(n/2)+3,....., an - 1, an twice.` ` ` `for` `(let i = 0; i < n/2; i++)` ` ` `{` ` ` `sum -= (2 * arr[i]);` ` ` `sum += (2 * arr[n - i - 1]);` ` ` `}` ` ` ` ` `return` `sum;` ` ` `}` `// Driver Code` ` ` `let arr = [ 4, 2, 1, 8 ];` ` ` `let n = arr.length;` ` ` `document.write(maxSum(arr, n));` `// This code is contributed by chinmoy1997pal.` `</script>` |

**Output : **

18

**Time Complexity: **O(nlogn). **Auxiliary Space :** O(1)

This article is contributed by **Anuj Chauhan**. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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