# Find a rotation with maximum hamming distance

Given an array of n elements, create a new array which is a rotation of given array and hamming distance between both the arrays is maximum.
Hamming distance between two arrays or strings of equal length is the number of positions at which the corresponding character(elements) are different.

Note: There can be more than one output for the given input.

Examples:

```Input :  1 4 1
Output :  2
Explanation:
Maximum hamming distance = 2.
We get this hamming distance with 4 1 1
or 1 1 4

input :  N = 4
2 4 8 0
output :  4
Explanation:
Maximum hamming distance = 4
We get this hamming distance with 4 8 0 2.
All the places can be occupied by another digit.
Other solutions can be 8 0 2 4, 4 0 2 8 etc.
```

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

Create another array which is double the size of the original array, such that the elements of this new array (copy array) are just the elements of the original array repeated twice in the same sequence. Example, if the original array is 1 4 1, then the copy array is 1 4 1 1 4 1.
Now, iterate through the copy array and find hamming distance with every shift (or rotation). So we check 4 1 1, 1 1 4, 1 4 1, choose the output for which the hamming distance is maximum.

Below is the implementation of above approach:

## C++

 `// C++ program to Find another array ` `// such that the hamming distance  ` `// from the original array is maximum ` `#include ` `using` `namespace` `std; ` ` `  `// Return the maximum hamming distance of a rotation ` `int` `maxHamming(``int` `arr[], ``int` `n) ` `{ ` `    ``// arr[] to brr[] two times so that ` `    ``// we can traverse through all rotations. ` `    ``int` `brr[2 *n + 1]; ` `    ``for` `(``int` `i = 0; i < n; i++) ` `        ``brr[i] = arr[i]; ` `    ``for` `(``int` `i = 0; i < n; i++)  ` `        ``brr[n+i] = arr[i]; ` ` `  `    ``// We know hamming distance with 0 rotation ` `    ``// would be 0. ` `    ``int` `maxHam = 0;     ` ` `  `    ``// We try other rotations one by one and compute ` `    ``// Hamming distance of every rotation ` `    ``for` `(``int` `i = 1; i < n; i++) ` `    ``{ ` `        ``int` `currHam = 0; ` `        ``for` `(``int` `j = i, k=0; j < (i + n); j++,k++)  ` `            ``if` `(brr[j] != arr[k]) ` `                 ``currHam++; ` ` `  `        ``// We can never get more than n.  ` `        ``if` `(currHam == n) ` `            ``return` `n; ` ` `  `        ``maxHam = max(maxHam, currHam); ` `    ``} ` ` `  `    ``return` `maxHam; ` `} ` ` `  `// driver program ` `int` `main()  ` `{ ` `    ``int` `arr[] = {2, 4, 6, 8};     ` `    ``int` `n = ``sizeof``(arr)/``sizeof``(arr); ` `    ``cout << maxHamming(arr, n);     ` `    ``return` `0; ` `} `

## Java

 `// Java program to Find another array ` `// such that the hamming distance  ` `// from the original array is maximum ` `class` `GFG  ` `{ ` `// Return the maximum hamming ` `// distance of a rotation ` `static` `int` `maxHamming(``int` `arr[], ``int` `n) ` `{ ` `    ``// arr[] to brr[] two times so that ` `    ``// we can traverse through all rotations. ` `    ``int` `brr[]=``new` `int``[``2` `*n + ``1``]; ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `        ``brr[i] = arr[i]; ` `    ``for` `(``int` `i = ``0``; i < n; i++)  ` `        ``brr[n+i] = arr[i]; ` ` `  `    ``// We know hamming distance with  ` `    ``// 0 rotation would be 0. ` `    ``int` `maxHam = ``0``;  ` ` `  `    ``// We try other rotations one by one  ` `    ``// and compute Hamming distance ` `    ``// of every rotation ` `    ``for` `(``int` `i = ``1``; i < n; i++) ` `    ``{ ` `        ``int` `currHam = ``0``; ` `        ``for` `(``int` `j = i, k=``0``; j < (i + n); j++, ` `                                          ``k++)  ` `            ``if` `(brr[j] != arr[k]) ` `                ``currHam++; ` ` `  `        ``// We can never get more than n.  ` `        ``if` `(currHam == n) ` `            ``return` `n; ` ` `  `        ``maxHam = Math.max(maxHam, currHam); ` `    ``} ` ` `  `    ``return` `maxHam; ` `}  ` ` `  `// driver code ` `public` `static` `void` `main (String[] args) ` `{ ` `    ``int` `arr[] = {``2``, ``4``, ``6``, ``8``};  ` `    ``int` `n = arr.length; ` `    ``System.out.print(maxHamming(arr, n));      ` `} ` `} ` ` `  `// This code is contributed by Anant Agarwal. `

## Python3

 `# Python3 code to Find another array ` `# such that the hamming distance  ` `# from the original array is maximum ` ` `  `# Return the maximum hamming  ` `# distance of a rotation ` `def` `maxHamming( arr , n ): ` ` `  `    ``# arr[] to brr[] two times so ` `    ``# that we can traverse through  ` `    ``# all rotations. ` `    ``brr ``=` `[``0``] ``*` `(``2` `*` `n ``+` `1``) ` `    ``for` `i ``in` `range``(n): ` `        ``brr[i] ``=` `arr[i] ` `    ``for` `i ``in` `range``(n): ` `        ``brr[n``+``i] ``=` `arr[i] ` `     `  `    ``# We know hamming distance  ` `    ``# with 0 rotation would be 0. ` `    ``maxHam ``=` `0` `     `  `    ``# We try other rotations one by ` `    ``# one and compute Hamming  ` `    ``# distance of every rotation ` `    ``for` `i ``in` `range``(``1``, n): ` `        ``currHam ``=` `0` `        ``k ``=` `0` `        ``for` `j ``in` `range``(i, i ``+` `n): ` `            ``if` `brr[j] !``=` `arr[k]: ` `                ``currHam ``+``=` `1` `                ``k ``=` `k ``+` `1` `         `  `        ``# We can never get more than n. ` `        ``if` `currHam ``=``=` `n: ` `            ``return` `n ` `         `  `        ``maxHam ``=` `max``(maxHam, currHam) ` `     `  `    ``return` `maxHam ` ` `  `# driver program ` `arr ``=` `[``2``, ``4``, ``6``, ``8``] ` `n ``=` `len``(arr) ` `print``(maxHamming(arr, n)) ` ` `  `# This code is contributed by "Sharad_Bhardwaj". `

## C#

 `// C# program to Find another array ` `// such that the hamming distance  ` `// from the original array is maximum ` `using` `System; ` ` `  `class` `GFG { ` `     `  `    ``// Return the maximum hamming ` `    ``// distance of a rotation ` `    ``static` `int` `maxHamming(``int` `[]arr, ``int` `n) ` `    ``{ ` `         `  `        ``// arr[] to brr[] two times so that ` `        ``// we can traverse through all rotations. ` `        ``int` `[]brr=``new` `int``[2 * n + 1]; ` `         `  `        ``for` `(``int` `i = 0; i < n; i++) ` `            ``brr[i] = arr[i]; ` `        ``for` `(``int` `i = 0; i < n; i++)  ` `            ``brr[n+i] = arr[i]; ` `     `  `        ``// We know hamming distance with  ` `        ``// 0 rotation would be 0. ` `        ``int` `maxHam = 0;  ` `     `  `        ``// We try other rotations one by one  ` `        ``// and compute Hamming distance ` `        ``// of every rotation ` `        ``for` `(``int` `i = 1; i < n; i++) ` `        ``{ ` `            ``int` `currHam = 0; ` `            ``for` `(``int` `j = i, k=0; j < (i + n);  ` `                                    ``j++, k++)  ` `                ``if` `(brr[j] != arr[k]) ` `                    ``currHam++; ` `     `  `            ``// We can never get more than n.  ` `            ``if` `(currHam == n) ` `                ``return` `n; ` `     `  `            ``maxHam = Math.Max(maxHam, currHam); ` `        ``} ` `     `  `        ``return` `maxHam; ` `    ``}  ` `     `  `    ``// driver code ` `    ``public` `static` `void` `Main () ` `    ``{ ` `        ``int` `[]arr = {2, 4, 6, 8};  ` `        ``int` `n = arr.Length; ` `        ``Console.Write(maxHamming(arr, n));  ` `    ``} ` `} ` ` `  `// This code is contributed by vt_m. `

## PHP

 ` `

Output:

```4
```

Time Complexity : O(n*n)

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