Skip to content
Related Articles

Related Articles

Minimize insertions in Array to divide it in pairs with Bitwise XOR as X

View Discussion
Improve Article
Save Article
  • Difficulty Level : Medium
  • Last Updated : 19 May, 2022
View Discussion
Improve Article
Save Article

Given an array arr of length N of distinct numbers and an integer X, the task is to find the minimum number of elements that should be added in the array such that the array elements can be grouped in pairs where the bitwise XOR of each pair is equal to X.

Examples:

 Input: N = 3, arr[] = {1, 2, 3}, X = 1
Output: 1
Explanation: If we add 0 to the array then the array becomes  {1, 2, 3, 0}. 
This array can be split as (1, 0), (2, 3), where XOR of each pair is 1.

Input: N = 5, arr[] = {1, 4, 5, 6, 7}, X = 7
Output: 3
Explanation: If we add (2, 0, 3) to the array, it becomes [1, 4, 5, 6, 7, 2, 0, 3]. 
This array can be split as (1, 6), (4, 3), (5, 2), (7, 0), with XOR of each pair as 7.

 

Approach: The problem can be solved using properties of Bitwise XOR operator:

According to question, let us assume that each pair of Array has bitwise XOR value as X, i.e.

arr[i] ⊕ arr[j] = X

Since above statement is true, therefore below statement will also be true

arr[i] ⊕ X = arr[j]

Based on above relation, we can solve this problem as following:

  • Find out existing pairs in Array with bitwise XOR value as X, and ignore them as they wont contribute to the solution.
  • From the remaining elements, the bitwise XOR of each element with X will be the element to be added in the Array to satisfy the given constraint.
  • Therefore count of remaining element will be the required count of insertions needed in the Array to split it into pairs with bitwise XOR as X.

Follow the steps mentioned below to implement the idea:

  • Create a frequency array to store the frequency of the array elements.
  • Insert the first element of the array in the hash data structure.
  • Run a loop from i = 1 to N-1
    • Check if A[i] ⊕ X  is already found in the array or not.
    • If not found then insert the current element in the hash data structure.
    • Otherwise, decrement the frequency of that element.
  • Return the total number of elements which has not yet formed a pair.

Below is the implementation of the above approach.

C++




// C++ code to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find out minimum number of
// element required to make array good
int minNumbersRequired(int N, int arr[], int X)
{
    // Variable to store the minimum number of
    // elements required
    int count = 0;
 
    // Map to store the frequency
    unordered_map<int, int> freq;
 
    freq[arr[0]]++;
    for (int i = 1; i < N; i++) {
 
        // If arr[i]^X is found,
        // then remove one occurrence
        // of both the elements from the map
        // as their bitwise XOR is already X
        if (freq[arr[i] ^ X] > 0) {
            freq[arr[i] ^ X]--;
        }
        else
            freq[arr[i]]++;
    }
 
    // Count the number of elements remaining
    // in the Array as the required answer
    for (auto it = freq.begin(); it != freq.end(); it++)
        count += it->second;
 
    // Return the minimised count of insertions
    return count;
}
 
// Driver Code
int main()
{
    int N = 5;
    int arr[] = { 1, 4, 5, 6, 7 };
    int X = 7;
 
    // Function call
    cout << minNumbersRequired(N, arr, X);
    return 0;
}


Java




// Java code to implement the approach
import java.io.*;
import java.util.*;
 
class GFG
{
   
    // Function to find out minimum number of
    // element required to make array good
    public static int minNumbersRequired(int N, int arr[],
                                         int X)
    {
        // Variable to store the minimum number of
        // elements required
        int count = 0;
 
        // Map to store the frequency
        HashMap<Integer, Integer> freq = new HashMap<>();
 
        freq.put(arr[0], 1);
        for (int i = 1; i < N; i++) {
 
            // If arr[i]^X is found,
            // then remove one occurrence
            // of both the elements from the map
            // as their bitwise XOR is already X
            if (freq.get(arr[i] ^ X) == null)
                freq.put(arr[i] ^ X, 1);
            else if (freq.get(arr[i] ^ X) > 0) {
                freq.put(arr[i] ^ X,
                         freq.get(arr[i] ^ X) - 1);
            }
            else
                freq.put(arr[i] ^ X, 1);
        }
 
        // Count the number of elements remaining
        // in the Array as the required answer
        for (Map.Entry<Integer, Integer> it :
             freq.entrySet())
            count += it.getValue();
 
        // Return the minimised count of insertions
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int N = 5;
        int arr[] = { 1, 4, 5, 6, 7 };
        int X = 7;
 
        // Function call
        System.out.print(minNumbersRequired(N, arr, X));
    }
}
 
// This code is contributed by Rohit Pradhan


Python3




# Python code to implement the approach
 
# Function to find out minimum number of
# element required to make array good
def minNumbersRequired(N, arr, X):
   
    # Variable to store the minimum number of
    # elements required
    count = 0
 
    # Map to store the frequency
    freq = {}
 
    freq[arr[0]] = 1
    for i in range(1,N):
 
        # If arr[i]^X is found,
        # then remove one occurrence
        # of both the elements from the map
        # as their bitwise XOR is already X
        if ((arr[i] ^ X) in freq and freq[arr[i] ^ X] > 0):
            freq[arr[i] ^ X] -= 1
        else:
            if (arr[i] in freq):
                freq[arr[i]] += 1
            else:
                freq[arr[i]] = 1
 
    # Count the number of elements remaining
    # in the Array as the required answer
    for it in freq:
        count += freq[it]
 
    # Return the minimised count of insertions
    return count
 
# Driver Code
N = 5
arr = [1, 4, 5, 6, 7]
X = 7
 
# Function call
print(minNumbersRequired(N, arr, X))
 
# This code is contributed by shinjanpatra


C#




// C# code to implement the approach
using System;
using System.Collections.Generic;
 
public class GFG
{
 
  // Function to find out minimum number of
  // element required to make array good
  public static int minNumbersRequired(int N, int[] arr,
                                       int X)
  {
 
    // Variable to store the minimum number of
    // elements required
    int count = 0;
 
    // Map to store the frequency
    Dictionary <int, int> freq = new Dictionary < int, int > ();
 
    freq[arr[0]] = 1;
    for (int i = 1; i < N; i++) {
 
      // If arr[i]^X is found,
      // then remove one occurrence
      // of both the elements from the map
      // as their bitwise XOR is already X
      if (!freq.ContainsKey(arr[i] ^ X))
        freq[arr[i] ^ X] = 1;
      else if (freq[arr[i] ^ X] > 0) {
        freq[arr[i] ^ X] -= 1;
      }
      else
        freq[arr[i] ^ X] = 1;
    }
 
    // Count the number of elements remaining
    // in the Array as the required answer
    foreach(KeyValuePair<int, int> it in freq)
      count += it.Value;
 
    // Return the minimised count of insertions
    return count;
  }
  public static void Main(string[] args)
  {
    int N = 5;
    int[] arr = { 1, 4, 5, 6, 7 };
    int X = 7;
 
    // Function call
    Console.WriteLine(minNumbersRequired(N, arr, X));
  }
}
 
// This code is contributed by phasing17


Javascript




<script>
    // JavaScript code to implement the approach
 
 
    // Function to find out minimum number of
    // element required to make array good
    const minNumbersRequired = (N, arr, X) => {
        // Variable to store the minimum number of
        // elements required
        let count = 0;
 
        // Map to store the frequency
        let freq = {};
 
        freq[arr[0]] = 1;
        for (let i = 1; i < N; i++) {
 
            // If arr[i]^X is found,
            // then remove one occurrence
            // of both the elements from the map
            // as their bitwise XOR is already X
            if ((arr[i] ^ X) in freq && freq[arr[i] ^ X] > 0) {
                freq[arr[i] ^ X]--;
            }
            else {
                if (arr[i] in freq)
                    freq[arr[i]]++;
                else freq[arr[i]] = 1;
            }
        }
 
        // Count the number of elements remaining
        // in the Array as the required answer
        for (let it in freq)
            count += freq[it];
 
        // Return the minimised count of insertions
        return count;
    }
 
    // Driver Code
 
    let N = 5;
    let arr = [1, 4, 5, 6, 7];
    let X = 7;
 
    // Function call
    document.write(minNumbersRequired(N, arr, X));
 
// This code is contributed by rakeshsahni
 
</script>


Output

3

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


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!