# Sort an Array based on the absolute difference of adjacent elements

Given an array arr[] containing N integers, the task is to rearrange all the elements of array such that absolute difference between consecutive elements of the array are sorted in increasing order.

Examples

Input: arr[] = { 5, -2, 4, 8, 6, 5 }
Output: 5 5 6 4 8 -2
Explanation:
|5 – 5| = 0
|5 – 6| = 1
|6 – 4| = 2
|4 – 8| = 4
|8 – (-2)| = 10
Hence, the differences between adjacent elements are sorted.

Input: arr[] = { 8, 1, 4, 2 }
Output: 4 2 8 1
Explanation:
|2 – 4| = 2
|8 – 2| = 6
|1 – 8| = 7
Hence, the differences between adjacent elements are sorted.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: The problem can be solved using Greedy Approach. We know that the maximum difference is between the minimum and maximum elements of the array. Using this fact, if we include one of the minimum element in the answer, then the next element included in the answer array will be the maximum element, then the third element included will be the second minimum, then the fourth element included will be the second maximum and so on will give the desired array.
Below are the steps:

1. Sort the given array arr[] in increasing order.
2. Choose the first maximum(say a) and minimum element(say b) from the sorted array and inserted in the answer array(say ans[]) as {a, b}.
3. Repeat the above steps by choosing the second, third, fourth… maximum and minimum element from the sorted array and insert it in the front of the answer array.
4. After all the above operations the answer array has the desired result.

Below is the implementation of the above approach:

 `// C++ implementation of the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function that arrange the array such that ` `// all absolute difference between adjacent ` `// element are sorted ` `void` `sortedAdjacentDifferences(``int` `arr[], ``int` `n) ` `{ ` `    ``// To store the resultant array ` `    ``int` `ans[n]; ` ` `  `    ``// Sorting the given array ` `    ``// in ascending order ` `    ``sort(arr + 0, arr + n); ` ` `  `    ``// Variable to represent left and right ` `    ``// ends of the given array ` `    ``int` `l = 0, r = n - 1; ` ` `  `    ``// Traversing the answer array in reverse ` `    ``// order and arrange the array elements from ` `    ``// arr[] in reverse order ` `    ``for` `(``int` `i = n - 1; i >= 0; i--) { ` ` `  `        ``// Inserting elements in zig-zag manner ` `        ``if` `(i % 2) { ` `            ``ans[i] = arr[l]; ` `            ``l++; ` `        ``} ` `        ``else` `{ ` `            ``ans[i] = arr[r]; ` `            ``r--; ` `        ``} ` `    ``} ` ` `  `    ``// Displaying the resultant array ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``cout << ans[i] << ``" "``; ` `    ``} ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 5, -2, 4, 8, 6, 4, 5 }; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` ` `  `    ``// Function Call ` `    ``sortedAdjacentDifferences(arr, n); ` `    ``return` `0; ` `} `

 `// Java implementation of the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` `  `  `// Function that arrange the array such that ` `// all absolute difference between adjacent ` `// element are sorted ` `static` `void` `sortedAdjacentDifferences(``int` `arr[], ``int` `n) ` `{ ` `    ``// To store the resultant array ` `    ``int` `[]ans = ``new` `int``[n]; ` `  `  `    ``// Sorting the given array ` `    ``// in ascending order ` `    ``Arrays.sort(arr); ` `  `  `    ``// Variable to represent left and right ` `    ``// ends of the given array ` `    ``int` `l = ``0``, r = n - ``1``; ` `  `  `    ``// Traversing the answer array in reverse ` `    ``// order and arrange the array elements from ` `    ``// arr[] in reverse order ` `    ``for` `(``int` `i = n - ``1``; i >= ``0``; i--) { ` `  `  `        ``// Inserting elements in zig-zag manner ` `        ``if` `(i % ``2` `== ``1``) { ` `            ``ans[i] = arr[l]; ` `            ``l++; ` `        ``} ` `        ``else` `{ ` `            ``ans[i] = arr[r]; ` `            ``r--; ` `        ``} ` `    ``} ` `  `  `    ``// Displaying the resultant array ` `    ``for` `(``int` `i = ``0``; i < n; i++) { ` `        ``System.out.print(ans[i]+ ``" "``); ` `    ``} ` `} ` `  `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``5``, -``2``, ``4``, ``8``, ``6``, ``4``, ``5` `}; ` `    ``int` `n = arr.length; ` `  `  `    ``// Function Call ` `    ``sortedAdjacentDifferences(arr, n); ` `} ` `} ` ` `  `// This code is contributed by Princi Singh `

 `# Python3 implementation of the above approach ` ` `  `# Function that arrange the array such that ` `# all absolute difference between adjacent ` `# element are sorted ` `def` `sortedAdjacentDifferences(arr, n): ` `     `  `    ``# To store the resultant array ` `    ``ans ``=` `[``0``]``*``n ` ` `  `    ``# Sorting the given array ` `    ``# in ascending order ` `    ``arr ``=` `sorted``(arr) ` ` `  `    ``# Variable to represent left and right ` `    ``# ends of the given array ` `    ``l ``=` `0` `    ``r ``=` `n ``-` `1` ` `  `    ``# Traversing the answer array in reverse ` `    ``# order and arrange the array elements from ` `    ``# arr[] in reverse order ` `    ``for` `i ``in` `range``(n ``-` `1``, ``-``1``, ``-``1``): ` ` `  `        ``# Inserting elements in zig-zag manner ` `        ``if` `(i ``%` `2``): ` `            ``ans[i] ``=` `arr[l] ` `            ``l ``+``=` `1` `        ``else``: ` `            ``ans[i] ``=` `arr[r] ` `            ``r ``-``=` `1` ` `  `    ``# Displaying the resultant array ` `    ``for` `i ``in` `range``(n): ` `        ``print``(ans[i], end``=``" "``) ` ` `  `# Driver Code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``arr``=``[``5``, ``-``2``, ``4``, ``8``, ``6``, ``4``, ``5``] ` `    ``n ``=` `len``(arr) ` ` `  `    ``# Function Call ` `    ``sortedAdjacentDifferences(arr, n) ` `     `  `# This code is contributed by mohit kumar 29 `

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` `    ``// Function that arrange the array such that ` `    ``// all absolute difference between adjacent ` `    ``// element are sorted ` `    ``static` `void` `sortedAdjacentDifferences(``int``[] arr, ``int` `n) ` `    ``{ ` `        ``// To store the resultant array ` `        ``int``[] ans = ``new` `int``[n]; ` `      `  `        ``// Sorting the given array ` `        ``// in ascending order ` `        ``Array.Sort(arr); ` `      `  `        ``// Variable to represent left and right ` `        ``// ends of the given array ` `        ``int` `l = 0, r = n - 1; ` `      `  `        ``// Traversing the answer array in reverse ` `        ``// order and arrange the array elements from ` `        ``// arr[] in reverse order ` `        ``for` `(``int` `i = n - 1; i >= 0; i--) { ` `      `  `            ``// Inserting elements in zig-zag manner ` `            ``if` `(i % 2 != 0) { ` `                ``ans[i] = arr[l]; ` `                ``l++; ` `            ``} ` `            ``else` `{ ` `                ``ans[i] = arr[r]; ` `                ``r--; ` `            ``} ` `        ``} ` `      `  `        ``// Displaying the resultant array ` `        ``for` `(``int` `i = 0; i < n; i++) { ` `            ``Console.Write(ans[i] + ``" "``); ` `        ``} ` `    ``} ` `      `  `    ``// Driver Code ` `    ``public` `static` `void` `Main() ` `    ``{ ` `        ``int``[] arr = { 5, -2, 4, 8, 6, 4, 5 }; ` `        ``int` `n = arr.Length; ` `      `  `        ``// Function Call ` `        ``sortedAdjacentDifferences(arr, n); ` `    ``} ` `} ` ` `  `// This code is contributed by chitranayal `

Output:
```5 4 5 4 6 -2 8
```

Time Complexity: O(N*log N), where N is the number of element in the given array.

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