Find a rotation with maximum hamming distance | Set 2

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Given an integer array arr[]. Create a new array which is a rotation of given array and find the maximum Hamming distance between both the arrays.

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.

Example:

Input: arr[] = [1, 4, 1]

Output: 2

Explanation: Maximum hamming distance = 2, this hamming distance with 4 1 1 or 1 1 4

Input: arr[] = [2, 4, 8, 0]

Output: 4

Explanation: Maximum hamming distance = 4, 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.

Approach: This problem can be solved efficiently by using the Hashmap. Follow the steps given below to solve the problem

Iterate through the array arr[] and store the values of the array in a HashMap <key, value>. Value can be List<> or string depending on interest.
Add the value in hashmap, hashMap.add(arrvalue, list).
Check this condition:
- When the hashMap.contains(arrValue) then hashmap.get(arrayValue).append(index).
Check the size of the hashmap, if size == 1 then
- max hamming distance = 0. Because all values are the same.
Run the for each loop, for each key if the list size = 1 then
- max hamming distance = n. Because all values are diff.
Create an array of size – 1 of the given array for storing the similar occurrences that can happen.
After storing all the indexes of array store the values in hashMap.
For each difference found a += 1. This difference says the number of rotations it would take to have the same value in the same position.
Find the least value in the array and get the index => min same values => max hamming distance.

Below is the implementation of the above approach:

C++

// Find a rotation with maximum hamming distance
#include<bits/stdc++.h>
using namespace std;
 
// Function to return the min value
int getMinValue(vector<int> numbers)
{
     
    // var to store the min value
    int minValue = numbers[0];
     
    for(int i = 1; i < numbers.size(); i++) 
    {
         
        // Condition to check if value is
        // lesser from the minValue or not
        if (numbers[i] < minValue) 
        {
            minValue = numbers[i];
        }
    }
     
    // Return the minValue
    return minValue;
}
 
// Function to get the hamming distance
int maxHammingDistance(vector<int> array)
{
    int n = array.size();
    vector<int> repetitionsOnRotations(n - 1,0);
     
    // Create the map to store the key value pair
    map<int, vector<int>> indexesOfElements;
     
    for(int i = 0; i < n; i++) 
    {
        int key = array[i];
         
        // Take empty list of integers
        // to store the integer
        vector<int>indexes;
         
        if (indexesOfElements.find(key) != 
            indexesOfElements.end())
        {
            indexes = indexesOfElements[key];
        } 
        else
        {
            indexes.clear();
        }
         
        // Add the value in list index
        indexes.push_back(i);
        indexesOfElements[key] = indexes;
    }
         
    // Run the for loop and get the
    // value from hash map one at a time
    for(auto x : indexesOfElements) 
    {
        vector<int> indexes = x.second;
         
        for(int i = 0; i < indexes.size() - 1; i++)
        {
            for(int j = i + 1; j < indexes.size(); j++) 
            {
                 
                // Store the diff into the variable
                int diff = indexes[i] - indexes[j];
                 
                // Check the condition if diff
                // is less than zero or not
                if (diff < 0) 
                {
                    repetitionsOnRotations[-diff - 1] += 1;
                    diff = n + diff;
                }
                repetitionsOnRotations += 1;
            }
        }
    }
     
    // Return the hamming distance
    return n - getMinValue(repetitionsOnRotations);
}
 
// Driver code
int main()
{
     
    // Example 1
    vector<int> array{ 1, 4, 1 };
    int result1 = maxHammingDistance(array);
    cout << result1 << endl;
     
    // Example 2
    vector<int> array2{ 2, 4, 8, 0 };
    int result2 = maxHammingDistance(array2);
    cout << result2 << endl;
}
 
// This code is contributed by ipg2016107

Java

// Find a rotation with maximum hamming distance
 
import java.io.*;
import java.util.*;
 
class GFG {
 
    // Function to return the min value
    public static int getMinValue(int[] numbers)
    {
 
        // var to store the min value
        int minValue = numbers[0];
 
        for (int i = 1; i < numbers.length; i++) {
 
            // Condition to check if value is
            // lesser from the minValue or not
            if (numbers[i] < minValue) {
                minValue = numbers[i];
            }
        }
 
        // Return the minValue
        return minValue;
    }
 
    // Function to get the hamming distance
    public static int maxHammingDistance(int[] array)
    {
 
        int n = array.length;
        int[] repetitionsOnRotations = new int[n - 1];
 
        // Create the map to store the key value pair
        Map<Integer, List<Integer> > indexesOfElements
            = new HashMap<>();
 
        for (int i = 0; i < n; i++) {
 
            int key = array[i];
 
            // Take empty list of integers
            // to store the integer
            List<Integer> indexes = null;
 
            if (indexesOfElements.containsKey(key)) {
                indexes = indexesOfElements.get(key);
            }
            else {
                indexes = new ArrayList<>();
            }
            // Add the value in list index
            indexes.add(i);
            indexesOfElements.put(key, indexes);
        }
 
        // Run the for loop and get the
        // value from hash map one at a time
        for (Map.Entry<Integer, List<Integer> > keys :
             indexesOfElements.entrySet()) {
 
            List<Integer> indexes = keys.getValue();
 
            for (int i = 0; i < indexes.size() - 1; i++) {
                for (int j = i + 1; j < indexes.size(); j++) {
 
                    // Store the diff into the variable
                    int diff
                        = indexes.get(i) - indexes.get(j);
 
                    // Check the condition if diff
                    // is less than zero or not
                    if (diff < 0) {
                        repetitionsOnRotations[(-diff) - 1] += 1;
                        diff = n + diff;
                    }
                    repetitionsOnRotations += 1;
                }
            }
        }
 
        // Return the hamming distance
        return n - (getMinValue(repetitionsOnRotations));
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        // Example 1
        int[] array = { 1, 4, 1 };
        int result1 = GFG.maxHammingDistance(array);
        System.out.println(result1);
 
        // Example 2
        int[] array2 = { 2, 4, 8, 0 };
        int result2 = GFG.maxHammingDistance(array2);
        System.out.println(result2);
    }
}

Python3

# Find a rotation with maximum hamming distance
 
# Function to return the min value
def getMinValue(numbers):
 
    # var to store the min value
    minValue = numbers[0]
 
    for i in range(1, len(numbers)):
 
        # Condition to check if value is
        # lesser from the minValue or not
        if (numbers[i] < minValue):
            minValue = numbers[i]
 
    # Return the minValue
    return minValue
 
 
# Function to get the hamming distance
def maxHammingDistance(array):
    n = len(array)
    repetitionsOnRotations = [0] * (n - 1)
 
    # Create the map to store the key value pair
    indexesOfElements = {}
 
    for i in range(n):
        key = array[i]
 
        # Take empty list of integers
        # to store the integer
        indexes = []
 
        if (key in indexesOfElements):
            indexes = indexesOfElements[key]
        else:
            indexes = []
 
        # Add the value in list index
        indexes.append(i)
        indexesOfElements[key] = indexes
 
    # Run the for loop and get the
    # value from hash map one at a time
    for indexes in indexesOfElements.values():
 
 
        for i in range(len(indexes) - 1):
            for j in range(i + 1, len(indexes)):
 
                # Store the diff into the variable
                diff = indexes[i] - indexes[j]
 
                # Check the condition if diff
                # is less than zero or not
                if (diff < 0):
                    repetitionsOnRotations[-diff - 1] += 1
                    diff = n + diff
 
                repetitionsOnRotations += 1
 
    # Return the hamming distance
    return n - getMinValue(repetitionsOnRotations)
 
 
# Driver code
 
# Example 1
array = [1, 4, 1]
result1 = maxHammingDistance(array)
print(result1)
 
# Example 2
array2 = [2, 4, 8, 0]
result2 = maxHammingDistance(array2)
print(result2)
 
# This code is contributed by _saurabh_jaiswal

C#

// Find a rotation with maximum hamming distance
using System;
using System.Collections.Generic;
 
class GFG{
 
// Function to return the min value
public static int getMinValue(int[] numbers)
{
     
    // var to store the min value
    int minValue = numbers[0];
 
    for(int i = 1; i < numbers.Length; i++)
    {
         
        // Condition to check if value is
        // lesser from the minValue or not
        if (numbers[i] < minValue)
        {
            minValue = numbers[i];
        }
    }
     
    // Return the minValue
    return minValue;
}
 
// Function to get the hamming distance
public static int maxHammingDistance(int[] array)
{
    int n = array.Length;
    int[] repetitionsOnRotations = new int[n - 1];
 
    // Create the map to store the key value pair
    Dictionary<int, 
          List<int>> indexesOfElements = new Dictionary<int, 
                                                   List<int>>();
 
    for(int i = 0; i < n; i++) 
    {
        int key = array[i];
 
        // Take empty list of integers
        // to store the integer
        List<int> indexes = null;
 
        if (indexesOfElements.ContainsKey(key)) 
        {
            indexes = indexesOfElements[key];
        }
        else
        {
            indexes = new List<int>();
        }
         
        // Add the value in list index
        indexes.Add(i);
        if (!indexesOfElements.ContainsKey(key)) 
            indexesOfElements.Add(key, indexes);
    }
 
    // Run the for loop and get the
    // value from hash map one at a time
    foreach(KeyValuePair<int, 
                    List<int>> keys in indexesOfElements) 
    {
        List<int> indexes = keys.Value;
 
        for(int i = 0; i < indexes.Count - 1; i++)
        {
            for(int j = i + 1; j < indexes.Count; j++) 
            {
                 
                // Store the diff into the variable
                int diff = indexes[i] - indexes[j];
 
                // Check the condition if diff
                // is less than zero or not
                if (diff < 0) 
                {
                    repetitionsOnRotations[(-diff) - 1] += 1;
                    diff = n + diff;
                }
                repetitionsOnRotations += 1;
            }
        }
    }
 
    // Return the hamming distance
    return n - (getMinValue(repetitionsOnRotations));
}
 
// Driver code
public static void Main(String[] args)
{
     
    // Example 1
    int[] array = { 1, 4, 1 };
    int result1 = GFG.maxHammingDistance(array);
    Console.WriteLine(result1);
 
    // Example 2
    int[] array2 = { 2, 4, 8, 0 };
    int result2 = GFG.maxHammingDistance(array2);
    Console.WriteLine(result2);
}
}
 
// This code is contributed by umadevi9616

Javascript

<script>
 
// Find a rotation with maximum hamming distance
 
// Function to return the min value
function getMinValue(numbers)
{
     
    // var to store the min value
    let minValue = numbers[0];
     
    for(let i = 1; i < numbers.length; i++) 
    {
         
        // Condition to check if value is
        // lesser from the minValue or not
        if (numbers[i] < minValue) 
        {
            minValue = numbers[i];
        }
    }
     
    // Return the minValue
    return minValue;
}
 
// Function to get the hamming distance
function maxHammingDistance(array)
{
    let n = array.length;
    let repetitionsOnRotations = new Array(n - 1).fill(0);
     
    // Create the map to store the key value pair
    let indexesOfElements = new Map();
     
    for(let i = 0; i < n; i++) 
    {
        let key = array[i];
         
        // Take empty list of integers
        // to store the integer
        let indexes = [];
         
        if (indexesOfElements.has(key))
        {
            indexes = indexesOfElements.get(key);
        } 
        else
        {
            indexes = [];
        }
         
        // Add the value in list index
        indexes.push(i);
        indexesOfElements.set(key, indexes);
    }
         
    // Run the for loop and get the
    // value from hash map one at a time
    for(let keys of indexesOfElements) 
    {
        let indexes = keys[1];
         
        for(let i = 0; i < indexes.length - 1; i++)
        {
            for(let j = i + 1; j < indexes.length; j++) 
            {
                 
                // Store the diff into the variable
                let diff = indexes[i] - indexes[j];
                 
                // Check the condition if diff
                // is less than zero or not
                if (diff < 0) 
                {
                    repetitionsOnRotations[-diff - 1] += 1;
                    diff = n + diff;
                }
                repetitionsOnRotations += 1;
            }
        }
    }
     
    // Return the hamming distance
    return n - getMinValue(repetitionsOnRotations);
}
 
// Driver code
 
// Example 1
let array = [ 1, 4, 1 ];
let result1 = maxHammingDistance(array);
document.write(result1 + "<br>");
 
// Example 2
let array2 = [ 2, 4, 8, 0 ];
let result2 = maxHammingDistance(array2);
document.write(result2);
 
// This code is contributed by gfgking
 
</script>

Output:

2
4

Time complexity: O(N)
Auxiliary Space: O(N)

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