Maximum possible XOR of every element in an array with another array

Two arrays A and B consisting of N elements are given. The task is to compute the maximum possible XOR of every element in array A with array B.

Examples:

Input :
A : 7 3 9 12
B : 1 3 5 2
Output : 6 6 12 15
Explanation : 1 xor 7 = 6, 5 xor 3 = 6, 5
xor 9 = 12, 3 xor 12 = 15.

A naive approach to solve this problem would be to check every element of array A with every other element of array B and print the maximum xor.

Time Complexity : O(n^2)

An efficient solution is to use Trie Data Structure.
We maintain a Trie for the binary representation of all elements in array B.
Now, for every element of A, we find the maximum xor in trie.
Let’s say our number A[i] is {b1, b2…bn}, where b1, b2…bn are binary bits. We start from b1.
Now for the xor to be maximum, we’ll try to make most significant bit 1 after performing
the xor. So, if b1 is 0, we’ll need a 1 and vice versa. In the trie, we go to the required
bit side. If favourable option is not there, we’ll go other side. Doing this all over,
we’ll get the maximum XOR possible.



Below is the implementation

C++

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// CPP code to find the maximum possible X-OR of
// every array element with another array
#include<bits/stdc++.h>
using namespace std;
  
// Structure of Trie DS
struct trie
{
    int value;
    trie *child[2];
};
  
// Function to initialise a Trie Node
trie * get()
{
    trie * root = new trie;
    root -> value = 0;
    root -> child[0] = NULL;
    root -> child[1] = NULL;
    return root;
}
  
// Computing max xor
int max_xor(trie * root, int key)
{
    trie * temp = root;
      
    // Checking for all bits in integer range
    for (int i = 31; i >= 0; i--)
    {
        // Current bit in the number
        bool current_bit = ( key & ( 1 << i) );
   
        // Traversing Trie for different bit, if found
        if (temp -> child[1 - current_bit] != NULL)
            temp = temp -> child[1 - current_bit];
   
        // Traversing Trie for same bit
        else
            temp = temp -> child[current_bit];
    }
   
    // Returning xor value of maximum bit difference 
    // value. Thus, we get maximum xor value
    return (key ^ temp -> value);
}
  
// Inserting B[] in Trie
void insert(trie * root, int key)
{
    trie * temp = root;
      
    // Storing 31 bits as integer representation
    for (int i = 31; i >= 0; i--)
    {
        // Current bit in the number
        bool current_bit = key & (1 << i);
          
        // New node required
        if (temp -> child[current_bit] == NULL)        
            temp -> child[current_bit] = get();
  
        // Traversing in Trie
        temp = temp -> child[current_bit];
    }
    // Assigning value to the leaf Node
    temp -> value = key;
}
  
int main()
{
    int A[] = {7, 3, 9, 12};
    int B[] = {1, 3, 5, 2};
      
    int N = sizeof(A)/sizeof(A[0]);
      
    // Forming Trie for B[]
    trie * root = get();
    for (int i = 0; i < N; i++)
        insert(root, B[i]);
      
    for (int i = 0; i < N; i++)
        cout << max_xor(root, A[i]) << endl;
      
    return 0;
}

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Java

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// Java code to find the maximum possible X-OR of
// every array element with another array
import java.util.*;
  
class GFG
{
  
// Structure of Trie DS
static class trie
{
    int value;
    trie []child = new trie[2];
};
  
// Function to initialise a Trie Node
static trie get()
{
    trie root = new trie();
    root.value = 0;
    root.child[0] = null;
    root.child[1] = null;
    return root;
}
  
// Computing max xor
static int max_xor(trie root, int key)
{
    trie temp = root;
      
    // Checking for all bits in integer range
    for (int i = 31; i >= 0; i--)
    {
        // Current bit in the number
        int current_bit = (key & (1 << i));
        if(current_bit > 0)
            current_bit = 1;
              
        // Traversing Trie for different bit, if found
        if (temp.child[1 - current_bit] != null)
            temp = temp.child[1 - current_bit];
  
        // Traversing Trie for same bit
        else
            temp = temp.child[current_bit];
    }
  
    // Returning xor value of maximum bit difference 
    // value. Thus, we get maximum xor value
    return (key ^ temp.value);
}
  
// Inserting B[] in Trie
static void insert(trie root, int key)
{
    trie temp = root;
      
    // Storing 31 bits as integer representation
    for (int i = 31; i >= 0; i--)
    {
        // Current bit in the number
        int current_bit = key & (1 << i);
        if(current_bit > 0)
            current_bit = 1;
              
        //System.out.println(current_bit);
        // New node required
        if (temp.child[current_bit] == null)     
            temp.child[current_bit] = get();
  
        // Traversing in Trie
        temp = temp.child[current_bit];
    }
    // Assigning value to the leaf Node
    temp.value = key;
}
  
// Driver Code
public static void main(String[] args)
{
    int A[] = {7, 3, 9, 12};
    int B[] = {1, 3, 5, 2};
      
    int N = A.length;
      
    // Forming Trie for B[]
    trie root = get();
    for (int i = 0; i < N; i++)
        insert(root, B[i]);
      
    for (int i = 0; i < N; i++)
        System.out.println(max_xor(root, A[i]));
}
}
  
// This code is contributed by 29AjayKumar

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C#

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// C# code to find the maximum possible X-OR of
// every array element with another array
using System;
      
class GFG
{
  
// Structure of Trie DS
class trie
{
    public int value;
    public trie []child = new trie[2];
};
  
// Function to initialise a Trie Node
static trie get()
{
    trie root = new trie();
    root.value = 0;
    root.child[0] = null;
    root.child[1] = null;
    return root;
}
  
// Computing max xor
static int max_xor(trie root, int key)
{
    trie temp = root;
      
    // Checking for all bits in integer range
    for (int i = 31; i >= 0; i--)
    {
        // Current bit in the number
        int current_bit = (key & (1 << i));
        if(current_bit > 0)
            current_bit = 1;
              
        // Traversing Trie for different bit, if found
        if (temp.child[1 - current_bit] != null)
            temp = temp.child[1 - current_bit];
  
        // Traversing Trie for same bit
        else
            temp = temp.child[current_bit];
    }
  
    // Returning xor value of maximum bit difference 
    // value. Thus, we get maximum xor value
    return (key ^ temp.value);
}
  
// Inserting B[] in Trie
static void insert(trie root, int key)
{
    trie temp = root;
      
    // Storing 31 bits as integer representation
    for (int i = 31; i >= 0; i--)
    {
        // Current bit in the number
        int current_bit = key & (1 << i);
        if(current_bit > 0)
            current_bit = 1;
              
        // System.out.println(current_bit);
        // New node required
        if (temp.child[current_bit] == null)     
            temp.child[current_bit] = get();
  
        // Traversing in Trie
        temp = temp.child[current_bit];
    }
      
    // Assigning value to the leaf Node
    temp.value = key;
}
  
// Driver Code
public static void Main(String[] args)
{
    int []A = {7, 3, 9, 12};
    int []B = {1, 3, 5, 2};
      
    int N = A.Length;
      
    // Forming Trie for B[]
    trie root = get();
    for (int i = 0; i < N; i++)
        insert(root, B[i]);
      
    for (int i = 0; i < N; i++)
        Console.WriteLine(max_xor(root, A[i]));
}
}
  
// This code is contributed by 29AjayKumar

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Output:

6
6
12
15

Trie formation : O(N x MAX_BITS) where MAX_BITS is maximum number of bits in binary representation of numbers.
Calculating max bit difference : O(N x MAX_BITS)
Time Complexity : O(N x MAX_BITS)

This article is contributed by Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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Improved By : 29AjayKumar

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