Minimum XOR Value Pair
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.
Implementation:
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 programint 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 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#
// 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.?> |
Javascript
<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.
Algorithm:
- Sort the given array
- 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 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;} |
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 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#
// 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 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.?> |
Javascript
<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)
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.
Algorithm:
- Create an empty trie. Every node of trie contains two children for 0 and 1 bits.
- Initialize min_xor = INT_MAX, insert arr[0] into trie
- Traversal all array element one-by-one starting from second.
- First find minimum setbet difference value in trie
- do xor of current element with minimum setbit diff that value
- update min_xor value if required
- insert current array element in trie
- 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 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
// 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 Nodeclass TrieNode: def __init__(self): # Child array with 0 and 1 self.child = [None]*2 # meant for the lead Node self.value = Noneclass 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 |
C#
// Include namespace systemusing System;// C# program to find minimum XOR value in an array.public class GFG{ public const int INT_SIZE = 32; // A Trie Node public class TrieNode { public int value; // used in leaf node public TrieNode[] Child = new TrieNode[2]; public TrieNode() { this.value = 0; this.Child[0] = null; this.Child[1] = null; } } public static TrieNode root; // utility function insert new key in trie public static void insert(int key) { var temp = root; // start from the most significant bit, insert all // bit of key one-by-one into trie for (int i = GFG.INT_SIZE - 1; i >= 0; i--) { // Find current bit in given prefix var 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. public static int minXORUtil(int key) { var temp = root; for (int i = GFG.INT_SIZE - 1; i >= 0; i--) { // Find current bit in given prefix var 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]; } 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] public static int minXOR(int[] arr, int n) { var min_xor = int.MaxValue; // Initialize result // create a True and insert first element in it root = new TrieNode(); GFG.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,GFG.minXORUtil(arr[i])); // insert current array value into Trie GFG.insert(arr[i]); } return min_xor; } // Driver code public static void Main(String[] args) { int[] arr = {9, 5, 3}; var n = arr.Length; Console.WriteLine(GFG.minXOR(arr, n)); }}// This code is contributed by aadityaburujwale. |
Javascript
class TrieNode {constructor() {this.child = new Array(2);this.value = null;}}class Trie {constructor() {this.root = this.getNode();}getNode() {return new TrieNode();}insert(key) {let temp = this.root;for (let i = 31; i >= 0; i--) {let curr = (key >> i) & 1;if (!temp.child[curr]) temp.child[curr] = this.getNode();temp = temp.child[curr];}temp.value = key;}xorUtil(key) {let temp = this.root;for (let i = 31; i >= 0; i--) {let curr = (key >> i) & 1;if (temp.child[curr]) temp = temp.child[curr];else if (temp.child[1 - curr]) temp = temp.child[1 - curr];}return temp.value ^ key;}}function minXor(arr) {let m = 2 ** 30;let trie = new Trie();trie.insert(arr[0]);for (let i = 1; i < arr.length; i++) {m = Math.min(m, trie.xorUtil(arr[i]));trie.insert(arr[i]);}return m;}if (typeof module !== 'undefined') {module.exports = {minXor: minXor,};}console.log(minXor([9, 5, 3]));// This code is contributed by akashish__ |
Output
6
Time Complexity O(n)
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