Minimum XOR Value Pair

3.6

Given an array of integers. Find the pair in an array which has minimum XOR value.
Examples:

Input : arr[] =  {9, 5, 3}
Output : 6
        All pair with xor value (9 ^ 5) => 12, 
        (5 ^ 3) => 6 , (9 ^ 3) => 10.
        Minimum XOR value is 6

Input : arr[] = {1, 2, 3, 4, 5}
Output : 1 

A Simple Solution is generate all pairs of given array and compute XOR their values. Finally return minimum XOR value. This solution takes O(n2) time.

C++

// C++ program to find minimum XOR value in an array.
#include<bits/stdc++.h>
using namespace std;

// Returns minimum xor value of pair in arr[0..n-1]
int minXOR(int arr[], int n)
{
    int min_xor = INT_MAX; // Initialize result

    // Generate all pair of given array
    for (int i=0; i < n; i++)
        for (int j=i+1; j<n; j++)

            // update minimum xor value if required
            min_xor = min(min_xor, arr[i] ^ arr[j] );

    return min_xor;
}

// Driver program
int main()
{
    int arr[] = {9 , 5 , 3};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << minXOR(arr, n) << endl;
    return 0;
}

Java

// Java program to find minimum XOR value in an array.
class GFG {

    // Returns minimum xor value of pair in arr[0..n-1]
    static int minXOR(int arr[], int n)
    {
        int min_xor = Integer.MAX_VALUE; // Initialize result
     
        // Generate all pair of given array
        for (int i=0; i < n; i++)
            for (int j=i+1; j<n; j++)
     
                // update minimum xor value if required
                min_xor = Math.min(min_xor, arr[i] ^ arr[j] );
     
        return min_xor;
    }
     
    // Driver program
    public static void main(String args[])
    {
        int arr[] = {9 , 5 , 3};
        int n = arr.length;
        System.out.println(minXOR(arr, n));
    }
}
//This code is contributed by Sumit Ghosh

An Efficient solution can solve this problem in O(nlogn) time. Below is the algorithm:

1). Sort the given array
2). Traverse and check XOR for every consecutive pair

Below is C++ implementation of above approach:

C++

#include<bits/stdc++.h>
using namespace std;

// Function to find minimum XOR pair
int minXOR(int arr[], int n)
{   
    // Sort given array
    sort(arr, arr+n);
    
    int minXor = INT_MAX;
    int val = 0;
    
    // calculate min xor of consecutive pairs
    for(int i=0;i<n-1;i++)
    {
        val = arr[i]^arr[i+1];
        minXor = min(minXor, val);
    }
    
    return minXor;
}

// Driver program
int main()
{
    int arr[] = {9 , 5 , 3};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << minXOR(arr, n) << endl;
    
    return 0;
}

Java

import java.util.Arrays;
class GFG {
    
    // Function to find minimum XOR pair
    static int minXOR(int arr[], int n)
    {   
        // Sort given array
        Arrays.parallelSort(arr);
         
        int minXor = Integer.MAX_VALUE;
        int val = 0;
         
        // calculate min xor of consecutive pairs
        for(int i=0;i<n-1;i++)
        {
            val = arr[i]^arr[i+1];
            minXor = Math.min(minXor, val);
        }
         
        return minXor;
    }
     
    // Driver program
    public static void main(String args[])
    {
        int arr[] = {9 , 5 , 3};
        int n = arr.length;
        System.out.println(minXOR(arr, n));
    }
}
// This code is contributed by Sumit Ghosh

Output:

6

Time Complexity: O(N*logN)
Space Complexity: O(1)
Thanks to Utkarsh Gupta for suggesting above approach.
A further more Efficient solution can solve the above problem in O(n) time under the assumption that integers take fixed number of bits to store. The idea is to use Trie Data Structure. Below is algorithm.

1). Create an empty trie. Every node of trie contains two children
    for 0 and 1 bits.
2). Initialize min_xor = INT_MAX, insert arr[0] into trie
3). Traversal all array element one-by-one starting from second.
     a. First find minimum setbet difference value in trie 
        do xor of current element with minimum setbit diff that value 
     b. update min_xor value if required
     c. insert current array element in trie 
4). return min_xor
  



Below is C++ implementation of above algorithm.

C++

// C++ program to find minimum XOR value in an array.
#include<bits/stdc++.h>
using namespace std;
#define INT_SIZE 32

// A Trie Node
struct TrieNode
{
    int value; // used in leaf node
    TrieNode * Child[2];
};

// Utility function to create a new Trie node
TrieNode * getNode()
{
    TrieNode * newNode = new TrieNode;
    newNode->value = 0;
    newNode->Child[0] = newNode->Child[1] = NULL;
    return newNode;
}

// utility function insert new key in trie
void insert(TrieNode *root, int key)
{
    TrieNode *temp = root;

    // start from the most significant bit , insert all
    // bit of key one-by-one into trie
    for (int i = INT_SIZE-1; i >= 0; i--)
    {
        // Find current bit in given prefix
        bool current_bit = (key & (1<<i));

        // Add a new Node into trie
        if (temp->Child[current_bit] == NULL)
            temp->Child[current_bit] = getNode();

        temp = temp->Child[current_bit];
    }

    // store value at leafNode
    temp->value = key ;
}

// Returns minimum XOR value of an integer inserted
// in Trie and given key.
int  minXORUtil(TrieNode * root, int key)
{
    TrieNode * temp = root;

    for (int i=INT_SIZE-1; i >= 0; i--)
    {
        // Find current bit in given prefix
        bool current_bit = ( key & ( 1<<i) );

        // Traversal Trie, look for prefix that has
        // same bit
        if (temp->Child[current_bit] != NULL)
            temp = temp->Child[current_bit];

        // if there is no same bit.then looking for
        // opposite bit
        else if(temp->Child[1-current_bit] !=NULL)
            temp = temp->Child[1-current_bit];
    }

    // return xor value of minimum bit difference value
    // so we get minimum xor value
    return key ^ temp->value;
}

// Returns minimum xor value of pair in arr[0..n-1]
int minXOR(int arr[], int n)
{
    int min_xor = INT_MAX;  // Initialize result

    // create a True and insert first element in it
    TrieNode *root = getNode();
    insert(root, arr[0]);

    // Traverse all array element and find minimum xor
    // for every element
    for (int i = 1 ; i < n; i++)
    {
        // Find minimum XOR value of current element with
        // previous elements inserted in Trie
        min_xor = min(min_xor, minXORUtil(root, arr[i]));

        // insert current array value into Trie
        insert(root, arr[i]);
    }
    return min_xor;
}

// Driver code
int main()
{
    int arr[] = {9, 5, 3};
    int n = sizeof(arr)/sizeof(arr[0]);
    cout << minXOR(arr, n) << endl;
    return 0;
}

Java

// Java program to find minimum XOR value in an array.
class GFG {
    static final int INT_SIZE = 32;
     
    // A Trie Node
    static class TrieNode
    {
        int value; // used in leaf node
        TrieNode[] Child = new TrieNode[2];
        
        public TrieNode() {
            value = 0;
            Child[0] = null;
            Child[1] = null;
        }
    }
    static TrieNode root;
    
    // utility function insert new key in trie
    static void insert(int key)
    {
        TrieNode temp = root;
     
        // start from the most significant bit , insert all
        // bit of key one-by-one into trie
        for (int i = INT_SIZE-1; i >= 0; i--)
        {
            // Find current bit in given prefix
            int current_bit = (key & (1<<i)) >= 1 ? 1 : 0;
            
            // Add a new Node into trie
            if (temp != null && temp.Child[current_bit] == null)
                temp.Child[current_bit] = new TrieNode();
     
            temp = temp.Child[current_bit];
        }
       
        // store value at leafNode
        temp.value = key ;
    }
     
    // Returns minimum XOR value of an integer inserted
    // in Trie and given key.
    static int  minXORUtil(int key)
    {
        TrieNode  temp = root;
     
        for (int i=INT_SIZE-1; i >= 0; i--)
        {
            // Find current bit in given prefix
            int current_bit = (key & (1<<i)) >= 1 ? 1 : 0;
                        
            // Traversal Trie, look for prefix that has
            // same bit
            if (temp.Child[current_bit] != null)
                temp = temp.Child[current_bit];
     
            // if there is no same bit.then looking for
            // opposite bit
            else if(temp.Child[1-current_bit] != null)
                temp = temp.Child[1-current_bit];
        }
        
        // return xor value of minimum bit difference value
        // so we get minimum xor value
        return key ^ temp.value;
    }
     
    // Returns minimum xor value of pair in arr[0..n-1]
    static int minXOR(int arr[], int n)
    {
        int min_xor = Integer.MAX_VALUE;  // Initialize result
     
        // create a True and insert first element in it
        root = new TrieNode();
        insert(arr[0]);
     
        // Traverse all array element and find minimum xor
        // for every element
        for (int i = 1 ; i < n; i++)
        {
            // Find minimum XOR value of current element with
            // previous elements inserted in Trie
            min_xor = Math.min(min_xor, minXORUtil(arr[i]));
     
            // insert current array value into Trie
            insert(arr[i]);
        }
        return min_xor;
    }
     
    // Driver code
    public static void main(String args[])
    {
        int arr[] = {9 , 5 , 3};
        int n = arr.length;
        System.out.println(minXOR(arr, n));
    }
}
// This code is contributed by Sumit Ghosh

Output:

6

Time Complexity O(n)

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