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 <bits/stdc++.h> 
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[0]); 
    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

<?php 
// PHP program to Find another array 
// such that the hamming distance  
// from the original array is maximum 
  
// Return the maximum hamming 
// distance of a rotation 
function maxHamming($arr, $n) 
{ 
      
    // arr[] to brr[] two times so that 
    // we can traverse through all rotations. 
    $brr = array(); 
    for ($i = 0; $i < $n; $i++) 
        $brr[$i] = $arr[$i]; 
    for ($i = 0; $i < $n; $i++)  
        $brr[$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 = 1; $i < $n; $i++) 
    { 
        $currHam = 0; 
        for ( $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 Code 
    $arr = array(2, 4, 6, 80);  
    $n = count($arr); 
    echo maxHamming($arr, $n);  
  
// This code is contributed by anuj_67. 
?> 

Output:

Time Complexity : O(n*n)

My Personal Notes

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

C++

Java

Python3

C#

PHP

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