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Minimum XOR-waardepaar

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Gegeven een array van gehele getallen. Zoek het paar in een array met een minimale XOR-waarde. 

Voorbeelden:  

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 Practice Minimaal XOR-waardepaar Probeer het!

A Eenvoudige oplossing is om alle paren van de gegeven array te genereren en XOR hun waarden te berekenen. Retourneer ten slotte de minimale XOR-waarde. Voor deze oplossing is O(n nodig) tijd. 



verander de mapnaam linux

Uitvoering:

C++
// C++ program to find minimum XOR value in an array. #include    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 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> 

Uitvoer
6

Ruimtecomplexiteit: O(1) 

Een Efficiënte oplossing kan dit probleem in O(nlogn) tijd oplossen. 

Algoritme: 

  1. Sorteer de gegeven array
  2. Doorloop en controleer XOR voor elk opeenvolgend paar

Hieronder vindt u de implementatie van bovenstaande aanpak:

C++
#include    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(0n-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
 // 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. ?> 
JavaScript
<script> // Function to find minimum XOR pair function 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> 

Uitvoer
6

Tijdcomplexiteit: O(N*logN) 
Ruimtecomplexiteit: O(1) 

Nog een verder Efficiënte oplossing kan het bovenstaande probleem in O(n) tijd oplossen onder de veronderstelling dat gehele getallen een vast aantal bits nodig hebben om op te slaan. Het idee is om Trie Data Structure te gebruiken.

Java-logo

Algoritme:

  1. Initialiseer min_xor = INT_MAX, voeg arr[0] in trie in
  2. Doorloop alle array-elementen één voor één, beginnend vanaf de tweede.
    1. Zoek eerst de minimale setbet-verschilwaarde in trie 
      • doe xor van huidig ​​element met minimale setbit diff die waarde 
    2. update de min_xor-waarde indien nodig
    3. voeg het huidige array-element in tri 
  3. retourneer min_xor

Hieronder vindt u de implementatie van het bovenstaande algoritme.

C++
// C++ program to find minimum XOR value in an array. #include    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 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(selfkey): 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(selfkey): 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(1len(arr)): # find the minimum xor value m = min(mtrie.xorUtil(arr[i])) # insert the new element trie.insert(arr[i]) return m # Driver Code  if __name__=='__main__': sample = [953] print(minXor(sample)) #code contributed by Shushant Kumar  
C#
// Include namespace system using 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_xorGFG.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__ 

Uitvoer
6

Tijdcomplexiteit O(n)

Ruimtecomplexiteit: O(n*INT_SIZE) 

 

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