Minimum XOR Value Pair - GeeksforGeeks

11 Mar, 2022
Given an array of integers. Find the pair in an array that has a 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 to generate all pairs of the given array and compute XOR their values. Finally, return minimum XOR value. This solution takes O(n2) time.  
C++
Java
Python3
C#
PHP
Javascript
// 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 programint main(){    int arr[] = { 9, 5, 3 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << minXOR(arr, n) << endl;    return 0;}
// 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
# Python program to find minimum# XOR value in an array. # Function to find minimum XOR pairdef minXOR(arr, n):         # Sort given array    arr.sort();     min_xor = 999999    val = 0     # calculate min xor of    # consecutive pairs    for i in range (0, n-1):        for j in range (i+1, n-1):                         # update minimum xor value            # if required            val = arr[i] ^ arr[j]            min_xor = min(min_xor, val)    return min_xor # Driver programarr = [ 9, 5, 3 ]n = len(arr) print(minXOR(arr, n)) # This code is contributed by Sam007.
// C# program to find minimum// XOR value in an array.using System; class GFG {         // Returns minimum xor value of    // pair in arr[0..n-1]    static int minXOR(int[] arr, int n)    {         // Initialize result        int min_xor = int.MaxValue;         // 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()    {        int[] arr = { 9, 5, 3 };        int n = arr.Length;        Console.WriteLine(minXOR(arr, n));    }} // This code is contributed by Sam007
<?php// PHP program to find minimum// XOR value in an array. // Returns minimum xor value// of pair in arr[0..n-1]function minXOR($arr, $n){    // Initialize result    $min_xor = PHP_INT_MAX;     // Generate all pair of given array    for ( $i = 0; $i < $n; $i++)        for ( $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 Code    $arr = array(9, 5, 3);    $n = count($arr);    echo minXOR($arr, $n);     // This code is contributed by anuj_67.?>
<script> // Javascript program to find// minimum XOR value in an array. // Returns minimum xor value of pair in arr[0..n-1]function minXOR(arr, n){    // Initialize result    let min_xor = Number.MAX_VALUE;     // Generate all pair of given array    for (let i = 0; i < n; i++)        for (let 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    let arr = [ 9, 5, 3 ];    let n = arr.length;    document.write(minXOR(arr, n)); </script>
Output:  
6
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 the implementation of above approach: 
C++
Java
Python3
C#
PHP
Javascript
#include <bits/stdc++.h>using namespace std; // Function to find minimum XOR pairint 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 programint main(){    int arr[] = { 9, 5, 3 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << minXOR(arr, n) << endl;     return 0;}
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
import sys    # Function to find minimum XOR pairdef minXOR(arr, n):         # Sort given array    arr.sort()      minXor =  int(sys.float_info.max)    val = 0      # calculate min xor of consecutive pairs    for i in range(0,n-1):        val = arr[i] ^ arr[i + 1];        minXor = min(minXor, val);         return minXor # Driver programarr = [9, 5, 3]n = len(arr)print(minXOR(arr, n))  # This code is contributed by Sam007.
// C# program to find minimum// XOR value in an array.using System; class GFG {         // Function to find minimum XOR pair    static int minXOR(int[] arr, int n)    {        // Sort given array        Array.Sort(arr);         int minXor = int.MaxValue;        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()    {        int[] arr = { 9, 5, 3 };        int n = arr.Length;        Console.WriteLine(minXOR(arr, n));    }} // This code is contributed by Sam007
<?php// Function to find minimum XOR pairfunction minXOR($arr, $n){    // Sort given array    sort($arr);     $minXor = PHP_INT_MAX;    $val = 0;     // calculate min xor    // of consecutive pairs    for ($i = 0; $i < $n - 1; $i++)    {        $val = $arr[$i] ^ $arr[$i + 1];        $minXor = min($minXor, $val);    }     return $minXor;} // Driver Code$arr = array(9, 5, 3);$n = count($arr);echo minXOR($arr, $n); // This code is contributed by Smitha.?>
<script> // Function to find minimum XOR pairfunction minXOR(arr, n){    // Sort given array    arr.sort();     let minXor = Number.MAX_VALUE;    let val = 0;     // calculate min xor of consecutive pairs    for (let i = 0; i < n - 1; i++) {        val = arr[i] ^ arr[i + 1];        minXor = Math.min(minXor, val);    }     return minXor;} // Driver program    let arr = [ 9, 5, 3 ];    let n = arr.length;    document.write(minXOR(arr, n)); </script>
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 the implementation of above algorithm. 
C++
Java
Python
// C++ program to find minimum XOR value in an array.#include <bits/stdc++.h>using namespace std;#define INT_SIZE 32 // A Trie Nodestruct TrieNode {    int value; // used in leaf node    TrieNode* Child[2];}; // Utility function to create a new Trie nodeTrieNode* getNode(){    TrieNode* newNode = new TrieNode;    newNode->value = 0;    newNode->Child[0] = newNode->Child[1] = NULL;    return newNode;} // utility function insert new key in trievoid 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 codeint main(){    int arr[] = { 9, 5, 3 };    int n = sizeof(arr) / sizeof(arr[0]);    cout << minXOR(arr, n) << endl;    return 0;}
// 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
# class for the basic Trie Nodeclass TrieNode:    def __init__(self):         # Child array with 0 and 1        self.child = [None]*2         # meant for the lead Node        self.value = None class Trie:     def __init__(self):        # initialise the root Node        self.root = self.getNode()     def getNode(self):        # get a new Trie Node        return TrieNode()     # inserts a new element    def insert(self,key):        temp = self.root          # 32 bit valued binary digit        for i in range(31,-1,-1):             # finding the bit at ith position            curr = (key>>i)&(1)             # if the child is None create one            if(temp.child[curr] is None):                temp.child[curr] = self.getNode()            temp = temp.child[curr]         # add value to the leaf node        temp.value = key     # traverse the trie and xor with the most similar element    def xorUtil(self,key):        temp = self.root         # 32 bit valued binary digit        for i in range(31,-1,-1):             # finding the bit at ith position            curr = (key>>i)&1             # traverse for the same bit            if(temp.child[curr] is not None):                temp = temp.child[curr]             # traverse if the same bit is not set in trie            else if(temp.child[1-curr] is not None):                temp = temp.child[1-curr]         # return with the xor of the value        return temp.value^key                      def minXor(arr):         # set m to a large number        m = 2**30         # initialize Trie        trie = Trie()         # insert the first element        trie.insert(arr[0])         # for each element in the array        for i in range(1,len(arr)):             # find the minimum xor value            m = min(m,trie.xorUtil(arr[i]))                         # insert the new element            trie.insert(arr[i])        return m # Driver Codeif __name__=="__main__":    sample = [9,5,3]    print(minXor(sample)) #code contributed by Ashwin Bhat   
Output:  
6
Time Complexity O(n)This article is contributed by Nishant_Singh (Pintu). If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
Sam007
vt_m
Smitha Dinesh Semwal
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