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 possible solutions are 8 0 2 4 and 0 2 4 8. Method #1:
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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 rotationint 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 programint 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 maximumclass GFG{// Return the maximum hamming// distance of a rotationstatic 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 codepublic 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 rotationdef 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 programarr = [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 maximumusing 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 rotationfunction 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.?> |
Javascript
<script>// JavaScript program to Find another array// such that the hamming distance// from the original array is maximum// Return the maximum hamming distance of a rotationfunction maxHamming(arr, n){ // arr[] to brr[] two times so that // we can traverse through all rotations. let brr = new Array(2 *n + 1); for (let i = 0; i < n; i++) brr[i] = arr[i]; for (let i = 0; i < n; i++) brr[n+i] = arr[i]; // We know hamming distance with 0 rotation // would be 0. let maxHam = 0; // We try other rotations one by one and compute // Hamming distance of every rotation for (let i = 1; i < n; i++) { let currHam = 0; for (let 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 let arr = [2, 4, 6, 8]; let n = arr.length; document.write(maxHamming(arr, n));// This code is contributed by Surbhi Tyagi.</script> |
Output:
4
Time Complexity : O(n*n)
Method #2:
We can find the maximum hamming distance using a different approach by taking advantage of list-comprehension in python. In this method, we divide the job in 3 separate functions.
- hamming_distance(x : list, y : list): This method returns the hamming distance for two list passed as parameters. The idea is to count the positions at which elements are different at the same index in two lists x and y where x is the original array taken in input and y is one of it rotations. Initialize a variable count from 0. Run loop from starting index 0 to last index (n-1) where n is the length of the list. For each iteration check if element of x and element at index i (0<=i<=n-1) is same or not. If they are same, increment the counter. After loop is completed, return the count(by definition this is the hamming distance for given arrays or strings)
- rotate_by_one(arr : list): This method rotates the array (passed in argument ) in anti-clockwise direction by 1 position. For e.g. if array [1,1,4,4] is passed, this method returns [1,4,4,5,1]. The idea is to copy the 1st element of the array and save it in a variable (say x). Then iterate the array from 0 to n-2 and copy every i+1 th value at ith position. Now assign x to last index.
- max_hamming_distance(arr : list): This method finds the maximum hamming distance for a given array and it’s rotations. Follow below steps in this method. We copy this array in a new array (say a) and initialize a variable max. Now, after every n rotations we get the original array. So we need to find hamming distance for original array with it’s n-1 rotations and store the current maximum in a variable(say max). Run loop for n-1 iterations. For each iteration, we follow below steps:
- Get the next rotation of arr by calling method ‘rotate_by_one’.
- Call method hamming distance() and pass original array (a) and current rotation of a (arr) and store the current hamming distance returned in a variable (say curr_h_dist).
- Check if value of curr_h_dist is greater than value of max. If yes, assign value of curr_h_dist to max_h.
- Repeat steps 1-3 till loop terminates.
- Return maximum hamming distance (max_h)
Python3
# Python code to to find maximum# of an array with it's rotationsimport time# Function hamming distance to find# the hamming distance for two lists/stringsdef hamming_distance(x: list, y: list): # Initialize count count = 0 # Run loop for size of x(or y) # as both as same length for i in range(len(x)): # Check if corresponding elements # at same index are not equal if(x[i] != y[i]): # Increment the count every # time above condition satisfies count += 1 # Return the hamming distance # for given pair of lists or strings return count# Function to rotate the given array# in anti-clockwise direction by 1def rotate_by_one(arr: list): # Store 1st element in a variable x = arr[0] # Update each ith element (0<=i<=n-2) # with it's next value for i in range(0, len(arr)-1): arr[i] = arr[i+1] # Assign 1st element to the last index arr[len(arr)-1] = x# Function max_hamming_distance to find# the maximum hamming distance for given arraydef max_hamming_distance(arr: list): # Initialize a variable to store # maximum hamming distance max_h = -10000000000 # Store size of the given array # in a variable n n = len(arr) # Initialize a new array a = [] # Copy the original array in new array for i in range(n): a.append(arr[i]) # Run loop for i=0 to i=n-1 for n-1 rotations for i in range(1, n): # Find the next rotation rotate_by_one(arr) print("Array after %d rotation : " % (i), arr) # Store hamming distance of current # rotation with original array curr_h_dist = hamming_distance(a, arr) print("Hamming Distance with %d rotations: %d" % (i, curr_h_dist)) # Check if current hamming distance # is greater than max hamming distance if curr_h_dist > max_h: # If yes, assign value of current # hamming distance to max hamming distance max_h = curr_h_dist print('\n') # Return maximum hamming distance return max_h# Driver codeif __name__ == '__main__': arr = [3, 0, 6, 4, 3] start = time.time() print('\n') print("Original Array : ", arr) print('\n') print("Maximum Hamming Distance: ", max_hamming_distance(arr)) end = time.time() print(f"Execution Time = {end - start}")# This code is contributed by Vivek_Kumar_Sinha |
Output
Original Array : [3, 0, 6, 4, 3] Array after 1 rotation : [0, 6, 4, 3, 3] Hamming Distance with 1 rotations: 4 Array after 2 rotation : [6, 4, 3, 3, 0] Hamming Distance with 2 rotations: 5 Array after 3 rotation : [4, 3, 3, 0, 6] Hamming Distance with 3 rotations: 5 Array after 4 rotation : [3, 3, 0, 6, 4] Hamming Distance with 4 rotations: 4 Maximum Hamming Distance: 5 Execution Time = 5.0067901611328125e-05
