Cuente los pares que tienen Bitwise XOR menos que K de una array dada

Posted on by Rudeus Greyrat

Dada una array arr[] de tamaño N y un número entero K , la tarea es contar el número de pares de la array dada de modo que el XOR bit a bit de cada par sea menor que K . 
 Ejemplos:

Entrada: arr = {1, 2, 3, 5} , K = 5 
Salida: 4 
Explicación: 
XOR bit a bit de todos los pares posibles que satisfacen las condiciones dadas son: 
arr[0] ^ arr[1] = 1 ^ 2 = 3 
arr[0] ^ arr[2] = 1 ^ 3 = 2 
arr[0] ^ arr[3] = 1 ^ 5 = 4 
arr[1] ^ arr[2] = 3 ^ 5 = 1 
Por lo tanto, la salida requerida es 4 

Entrada: arr[] = {3, 5, 6, 8}, K = 7 
Salida: 3  

Enfoque ingenuo: el enfoque más simple para resolver este problema es atravesar la array dada y generar todos los pares posibles de la array dada y para cada par, verificar si el XOR bit a bit del par es menor que K o no. Si se determina que es cierto, incremente el recuento de pares que tengan XOR bit a bit menor que K . Finalmente, imprima el conteo de dichos pares obtenidos.

Tiempo Complejidad :O(N 2 )
Espacio Auxiliar: O(1)

Enfoque eficiente: el problema se puede resolver usando Trie . La idea es iterar sobre la array dada y para cada elemento de la array, contar la cantidad de elementos presentes en el Trie cuyo XOR bit a bit con el elemento actual es menor que K e insertar la representación binaria del elemento actual en el Trie . Finalmente, imprima el conteo de pares que tienen XOR bit a bit menor que K . Siga los pasos a continuación para resolver el problema:

  • Cree un Trie que tenga un Node raíz, digamos raíz para almacenar la representación binaria de cada elemento de la array dada.
  • Recorra la array dada y cuente el número de elementos presentes en el Trie cuyo XOR bit a bit con el elemento actual es menor que K e inserte la representación binaria del elemento actual .
  • Finalmente, imprime el conteo de pares que satisface la condición dada.

A continuación se muestra la implementación del enfoque anterior:

C++

// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Structure of Trie
struct TrieNode
{
    // Stores binary representation
    // of numbers
    TrieNode *child[2];
 
    // Stores count of elements
    // present in a node
    int cnt;
     
    // Function to initialize
    // a Trie Node
    TrieNode() {
        child[0] = child[1] = NULL;
        cnt = 0;
    }
};
 
 
// Function to insert a number into Trie
void insertTrie(TrieNode *root, int N) {
     
    // Traverse binary representation of X
    for (int i = 31; i >= 0; i--) {
         
        // Stores ith bit of N
        bool x = (N) & (1 << i);
         
        // Check if an element already
        // present in Trie having ith bit x
        if(!root->child[x]) {
             
            // Create a new node of Trie.
            root->child[x] = new TrieNode();
        }
         
        // Update count of elements
        // whose ith bit is x
        root->child[x]->cnt+= 1;
         
        // Update root
        root= root->child[x];
    }
}
 
 
// Function to count elements
// in Trie whose XOR with N
// less than K
int cntSmaller(TrieNode * root,
                int N, int K)
{
     
    // Stores count of elements
    // whose XOR with N less than K
    int cntPairs = 0;
     
    // Traverse binary representation
    // of N and K in Trie
    for (int i = 31; i >= 0 &&
                    root; i--) {
                                     
        // Stores ith bit of N                        
        bool x = N & (1 << i);
         
        // Stores ith bit of K
        bool y = K & (1 << i);
         
        // If the ith bit of K is 1
        if (y) {
             
            // If an element already
            // present in Trie having
            // ith bit (x)
            if(root->child[x]) {
                    cntPairs  +=
                    root->child[x]->cnt;
            }
         
            root =
                root->child[1 - x];
        }
         
        // If the ith bit of K is 0
        else{
             
            // Update root
            root = root->child[x];
        }
    }
    return cntPairs;
}
 
// Function to count pairs that
// satisfy the given conditions
int cntSmallerPairs(int arr[], int N,
                            int K) {
     
    // Create root node of Trie
    TrieNode *root = new TrieNode();
     
    // Stores count of pairs that
    // satisfy the given conditions
    int cntPairs = 0;
     
    // Traverse the given array
    for(int i = 0;i < N; i++){
         
        // Update cntPairs
        cntPairs += cntSmaller(root,
                        arr[i], K);
         
        // Insert arr[i] into Trie
        insertTrie(root, arr[i]);
    }
    return cntPairs;
}
 
// Driver Code
int main()
{
    int arr[] = {3, 5, 6, 8};
    int K= 7;
    int N = sizeof(arr) / sizeof(arr[0]);
     
    cout<<cntSmallerPairs(arr, N, K);
}

Java

// Java program to implement
// the above approach
import java.util.*;
class GFG{
 
// Structure of Trie
static class TrieNode
{
  // Stores binary representation
  // of numbers
  TrieNode child[] = new TrieNode[2];
 
  // Stores count of elements
  // present in a node
  int cnt;
 
  // Function to initialize
  // a Trie Node
  TrieNode()
  {
    child[0] = child[1] = null;
    cnt = 0;
  }
};
 
 
// Function to insert a number
// into Trie
static void insertTrie(TrieNode root,
                       int N)
{   
  // Traverse binary representation
  // of X
  for (int i = 31; i >= 0; i--)
  {
    // Stores ith bit of N
    int x = (N) & (1 << i);
 
    // Check if an element already
    // present in Trie having ith
    // bit x
    if(x <2 && root.child[x] ==
       null)
    {
      // Create a new node of
      // Trie.
      root.child[x] = new TrieNode();
    }
 
    // Update count of elements
    // whose ith bit is x
    if(x < 2)
      root.child[x].cnt += 1;
 
    // Update root
    if(x < 2)
      root = root.child[x];
  }
}
 
// Function to count elements
// in Trie whose XOR with N
// less than K
static int cntSmaller(TrieNode root,
                      int N, int K)
{    
  // Stores count of elements
  // whose XOR with N less
  // than K
  int cntPairs = 0;
 
  // Traverse binary
  // representation of N
  // and K in Trie
  for (int i = 31; i >= 0 &&
       root != null; i--)
  {
    // Stores ith bit of N                        
    int x = (N & (1 << i));
 
    // Stores ith bit of K
    int y = (K & (1 << i));
 
    // If the ith bit of K
    // is 1
    if (y == 1)
    {
      // If an element already
      // present in Trie having
      // ith bit (x)
      if(root.child[x] != null)
      {
        cntPairs +=
           root.child[x].cnt;
      }
 
      root = root.child[1 - x];
    }
 
    // If the ith bit of K is 0
    else
    {
      // Update root
      if(x < 2)
        root = root.child[x];
    }
  }
  return cntPairs;
}
 
// Function to count pairs that
// satisfy the given conditions
static int cntSmallerPairs(int arr[],
                           int N, int K)
{   
  // Create root node of Trie
  TrieNode root = new TrieNode();
 
  // Stores count of pairs that
  // satisfy the given conditions
  int cntPairs = 0;
 
  // Traverse the given array
  for(int i = 0; i < N; i++)
  {
    // Update cntPairs
    cntPairs += cntSmaller(root,
                           arr[i], K);
 
    // Insert arr[i] into Trie
    insertTrie(root, arr[i]);
  }
  return cntPairs;
}
 
// Driver Code
public static void main(String[] args)
{
  int arr[] = {3, 5, 6, 8};
  int K= 7;
  int N = arr.length;
  System.out.print(cntSmallerPairs(arr,
                                   N, K));
}
}
 
// This code is contributed by Rajput-Ji

Python3

# Python3 program to implement
# the above approach
  
# Structure of Trie
class TrieNode:
 
    # Function to initialize
    # a Trie Node
    def __init__(self):       
        self.child = [None, None]
        self.cnt = 0
 
# Function to insert a number into Trie
def insertTrie(root, N):
      
    # Traverse binary representation of X.
    for i in range(31, -1, -1):
              
        # Stores ith bit of N
        x = bool( (N) & (1 << i));
          
        # Check if an element already
        # present in Trie having ith bit x.
        if(root.child[x] == None):
              
            # Create a new node of Trie.
            root.child[x] = TrieNode();
                  
        # Update count of elements
        # whose ith bit is x
        root.child[x].cnt += 1;
          
        # Update root.
        root= root.child[x];
   
# Function to count elements
# in Trie whose XOR with N
# less than K
def cntSmaller(root, N, K):
      
    # Stores count of elements
    # whose XOR with N exceeding K
    cntPairs = 0;
      
    # Traverse binary representation
    # of N and K in Trie
    for i in range(31, -1, -1):      
        if(root == None):
            break
                                     
        # Stores ith bit of N                        
        x = bool(N & (1 << i))
          
        # Stores ith bit of K
        y = K & (1 << i);
          
        # If the ith bit of K is 1
        if (y != 0):
             
            # If an element already
            # present in Trie having
            # ith bit (1 - x)
            if (root.child[x]):
 
                # Update cntPairs
                cntPairs += root.child[ x].cnt
 
            # Update root.
            root = root.child[1 - x];
          
        # If the ith bit of K is 0
        else:
              
            # Update root.
            root = root.child[x]
    return cntPairs;
  
# Function to count pairs that
# satisfy the given conditions.
def cntSmallerPairs(arr, N, K):
      
    # Create root node of Trie
    root = TrieNode();
      
    # Stores count of pairs that
    # satisfy the given conditions
    cntPairs = 0;
      
    # Traverse the given array.
    for i in range(N):
          
        # Update cntPairs
        cntPairs += cntSmaller(root, arr[i], K);
          
        # Insert arr[i] into Trie.
        insertTrie(root, arr[i]);   
    return cntPairs;
  
# Driver code
if __name__=='__main__':
    arr = [3, 5, 6, 8]
    K= 7;
    N = len(arr)    
    print(cntSmallerPairs(arr, N, K))
 
    # This code is contributed by rutvik_56

C#

// C# program to implement
// the above approach
using System;
 
class GFG{
 
// Structure of Trie
public class TrieNode
{
   
  // Stores binary representation
  // of numbers
  public TrieNode []child = new TrieNode[2];
 
  // Stores count of elements
  // present in a node
  public int cnt;
 
  // Function to initialize
  // a Trie Node
  public TrieNode()
  {
    child[0] = child[1] = null;
    cnt = 0;
  }
};
 
// Function to insert a number
// into Trie
static void insertTrie(TrieNode root,
                       int N)
{
   
  // Traverse binary representation
  // of X
  for(int i = 31; i >= 0; i--)
  {
     
    // Stores ith bit of N
    int x = (N) & (1 << i);
 
    // Check if an element already
    // present in Trie having ith
    // bit x
    if (x < 2 && root.child[x] == null)
    {
       
      // Create a new node of
      // Trie.
      root.child[x] = new TrieNode();
    }
 
    // Update count of elements
    // whose ith bit is x
    if (x < 2)
      root.child[x].cnt += 1;
 
    // Update root
    if (x < 2)
      root = root.child[x];
  }
}
 
// Function to count elements
// in Trie whose XOR with N
// less than K
static int cntSmaller(TrieNode root,
                      int N, int K)
{   
   
  // Stores count of elements
  // whose XOR with N less
  // than K
  int cntPairs = 0;
 
  // Traverse binary
  // representation of N
  // and K in Trie
  for(int i = 31; i >= 0 &&
      root != null; i--)
  {
     
    // Stores ith bit of N                        
    int x = (N & (1 << i));
 
    // Stores ith bit of K
    int y = (K & (1 << i));
 
    // If the ith bit of K
    // is 1
    if (y == 1)
    {
       
      // If an element already
      // present in Trie having
      // ith bit (x)
      if (root.child[x] != null)
      {
        cntPairs += root.child[x].cnt;
      }
       
      root = root.child[1 - x];
    }
 
    // If the ith bit of K is 0
    else
    {
       
      // Update root
      if (x < 2)
        root = root.child[x];
    }
  }
  return cntPairs;
}
 
// Function to count pairs that
// satisfy the given conditions
static int cntSmallerPairs(int []arr,
                           int N, int K)
{   
   
  // Create root node of Trie
  TrieNode root = new TrieNode();
 
  // Stores count of pairs that
  // satisfy the given conditions
  int cntPairs = 0;
 
  // Traverse the given array
  for(int i = 0; i < N; i++)
  {
     
    // Update cntPairs
    cntPairs += cntSmaller(root,
                           arr[i], K);
 
    // Insert arr[i] into Trie
    insertTrie(root, arr[i]);
  }
  return cntPairs;
}
 
// Driver Code
public static void Main(String[] args)
{
  int []arr = { 3, 5, 6, 8 };
  int K= 7;
  int N = arr.Length;
   
  Console.Write(cntSmallerPairs(arr,
                                N, K));
}
}
 
// This code is contributed by Princi Singh
Producción: 

3

 

Tiempo Complejidad :O(N * 32)
Espacio Auxiliar: O(N * 32)

Publicación traducida automáticamente

Artículo escrito por doreamon_ y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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