Minimum XOR Value Pair - GeeksforGeeks

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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 

Python3

# Python program to find minimum 
# XOR value in an array. 
  
# Function to find minimum XOR pair 
def 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 program 
arr = [ 9, 5, 3 ] 
n = len(arr) 
  
print(minXOR(arr, n)) 
  
# This code is contributed by Sam007. 

C#

// 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 
// 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. 
?> 

Output :

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++

#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 

Python3

import sys     
  
# Function to find minimum XOR pair 
def 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 program 
arr = [9, 5, 3] 
n = len(arr) 
print(minXOR(arr, n)) 
   
# This code is contributed by Sam007. 

C#

// 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

<?php 
// Function to find minimum XOR pair 
function 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. 
?> 

Output :

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++

// 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 

Python

# class for the basic Trie Node  
class TrieNode: 
    def __init__(self): 
  
        # Child array with 0 and 1  
        self.child = [None]*2 
  
        # meant foor 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 
            elif(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 Code  
if __name__=="__main__": 
    sample = [9,5,3] 
    print(minXor(sample)) 
  
#code contributed by Ashwin Bhat     

Output :

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 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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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

C++

Java

Python

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