# Minimum XOR Value Pair

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
```

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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