Minimum XOR Value Pair

http://www.geeksforgeeks.org/minimum-xor-value-pair/
Given an array of integers. Find the pair in an array which has minimum XOR value

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;

}

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

Time Complexity: O(N*logN)

    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;

    }

    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;

    }

An 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

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]);

    // Traversr all array elementd 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;

}