Suma de XOR de todos los subconjuntos de longitud K

Dada una array de longitud n (n > k), tenemos que encontrar la suma de xor de todos los elementos de las sub-arrays que son de longitud k.
Ejemplos: 
 

Entrada: arr[]={1, 2, 3, 4}, k=2 
Salida: Suma= 11 
Suma = 1^2 + 2^3 + 3^4 = 3 + 1 + 7 =11
Entrada: arr[] ={1, 2, 3, 4}, k=3 
Salida: Suma= 5 
Suma = 1^2^3 + 2^3^4 = 0 + 5 =5 
 

Solución ingenua: la idea es atravesar todos los subarreglos de longitud k y encontrar el xor de todos los elementos del subarreglo y sumarlos para encontrar la suma de XOR de todos los subarreglos de longitud K de un arreglo. 
Complejidad de tiempo : O(N 2 )
Solución eficiente: La solución eficiente es recorrer el arreglo y encontrar todos los subarreglo de longitud k, es decir (0 a k-1), (1 a k), (2 a k+1) , …., (n-k+1 a n).
Encontraremos y almacenaremos el xor de los elementos de 0 a i (en una array x[]) formando una array pre-xor.
Ahora, xor del subarreglo de l a r es igual a x[l-1] ^ x[r] porque x[r] dará el xor de todos los elementos hasta r y x[l-1] dará el xor de todos los elementos hasta l-1. Cuando tomemos xor de estos dos valores se repetirán los elementos hasta 0 a l-1. Como a^a = 0, los valores repetidos contribuirían con cero al valor neto y obtenemos el valor de xor subarray de l a r.
A continuación se muestra la implementación del enfoque anterior: 
 

C++

// C++ implementation of above approach
#include <iostream>
using namespace std;
 
// Sum of XOR of all K length
// sub-array of an array
int FindXorSum(int arr[], int k, int n)
{
    // If the length of the array is less than k
    if (n < k)
        return 0;
 
    // Array that will store xor values of
    // subarray from 1 to i
    int x[n] = { 0 };
    int result = 0;
 
    // Traverse through the array
    for (int i = 0; i < n; i++) {
 
        // If i is greater than zero, store
        // xor of all the elements from 0 to i
        if (i > 0)
            x[i] = x[i - 1] ^ arr[i];
 
        // If it is the first element
        else
            x[i] = arr[i];
 
        // If i is greater than k
        if (i >= k - 1) {
            int sum = 0;
 
            // Xor of values from 0 to i
            sum = x[i];
 
            // Now to find subarray of length k
            // that ends at i, xor sum with x[i-k]
            if (i - k > -1)
                sum ^= x[i - k];
 
            // Add the xor of elements from i-k+1 to i
            result += sum;
        }
    }
 
    // Return the resultant sum;
    return result;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
 
    int n = 4, k = 2;
 
    cout << FindXorSum(arr, k, n) << endl;
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
// Sum of XOR of all K length
// sub-array of an array
static int FindXorSum(int arr[], int k, int n)
{
    // If the length of the array is less than k
    if (n < k)
        return 0;
 
    // Array that will store xor values of
    // subarray from 1 to i
    int []x = new int[n];
    int result = 0;
 
    // Traverse through the array
    for (int i = 0; i < n; i++)
    {
 
        // If i is greater than zero, store
        // xor of all the elements from 0 to i
        if (i > 0)
            x[i] = x[i - 1] ^ arr[i];
 
        // If it is the first element
        else
            x[i] = arr[i];
 
        // If i is greater than k
        if (i >= k - 1)
        {
            int sum = 0;
 
            // Xor of values from 0 to i
            sum = x[i];
 
            // Now to find subarray of length k
            // that ends at i, xor sum with x[i-k]
            if (i - k > -1)
                sum ^= x[i - k];
 
            // Add the xor of elements from i-k+1 to i
            result += sum;
        }
    }
 
    // Return the resultant sum;
    return result;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4 };
 
    int n = 4, k = 2;
 
    System.out.println(FindXorSum(arr, k, n));
}
}
 
// This code contributed by Rajput-Ji

Python3

# Python implementation of above approach
 
# Sum of XOR of all K length
# sub-array of an array
def FindXorSum(arr, k, n):
     
    # If the length of the array is less than k
    if (n < k):
        return 0;
 
    # Array that will store xor values of
    # subarray from 1 to i
    x = [0]*n;
    result = 0;
 
    # Traverse through the array
    for i in range(n):
 
        # If i is greater than zero, store
        # xor of all the elements from 0 to i
        if (i > 0):
            x[i] = x[i - 1] ^ arr[i];
 
        # If it is the first element
        else:
            x[i] = arr[i];
 
        # If i is greater than k
        if (i >= k - 1):
            sum = 0;
 
            # Xor of values from 0 to i
            sum = x[i];
 
            # Now to find subarray of length k
            # that ends at i, xor sum with x[i-k]
            if (i - k > -1):
                sum ^= x[i - k];
 
            # Add the xor of elements from i-k+1 to i
            result += sum;
 
    # Return the resultant sum;
    return result;
 
# Driver code
arr = [ 1, 2, 3, 4 ];
 
n = 4; k = 2;
 
print(FindXorSum(arr, k, n));
 
# This code has been contributed by 29AjayKumar

C#

// C# implementation of the above approach
using System;
 
class GFG
{
     
    // Sum of XOR of all K length
    // sub-array of an array
    static int FindXorSum(int []arr, int k, int n)
    {
        // If the length of the array is less than k
        if (n < k)
            return 0;
     
        // Array that will store xor values of
        // subarray from 1 to i
        int []x = new int[n];
        int result = 0;
     
        // Traverse through the array
        for (int i = 0; i < n; i++)
        {
     
            // If i is greater than zero, store
            // xor of all the elements from 0 to i
            if (i > 0)
                x[i] = x[i - 1] ^ arr[i];
     
            // If it is the first element
            else
                x[i] = arr[i];
     
            // If i is greater than k
            if (i >= k - 1)
            {
                int sum = 0;
     
                // Xor of values from 0 to i
                sum = x[i];
     
                // Now to find subarray of length k
                // that ends at i, xor sum with x[i-k]
                if (i - k > -1)
                    sum ^= x[i - k];
     
                // Add the xor of elements from i-k+1 to i
                result += sum;
            }
        }
     
        // Return the resultant sum;
        return result;
    }
 
    // Driver code
    public static void Main()
    {
        int []arr = { 1, 2, 3, 4 };
     
        int n = 4, k = 2;
     
        Console.WriteLine(FindXorSum(arr, k, n));
    }
}
 
// This code is contributed by AnkitRai01

Javascript

<script>
 
// Javascript implementation of above approach
 
// Sum of XOR of all K length
// sub-array of an array
function FindXorSum(arr, k, n)
{
    // If the length of the array is less than k
    if (n < k)
        return 0;
 
    // Array that will store xor values of
    // subarray from 1 to i
    let x = new Array(n).fill(0);
    let result = 0;
 
    // Traverse through the array
    for (let i = 0; i < n; i++) {
 
        // If i is greater than zero, store
        // xor of all the elements from 0 to i
        if (i > 0)
            x[i] = x[i - 1] ^ arr[i];
 
        // If it is the first element
        else
            x[i] = arr[i];
 
        // If i is greater than k
        if (i >= k - 1) {
            let sum = 0;
 
            // Xor of values from 0 to i
            sum = x[i];
 
            // Now to find subarray of length k
            // that ends at i, xor sum with x[i-k]
            if (i - k > -1)
                sum ^= x[i - k];
 
            // Add the xor of elements from i-k+1 to i
            result += sum;
        }
    }
 
    // Return the resultant sum;
    return result;
}
 
// Driver code
    let arr = [ 1, 2, 3, 4 ];
 
    let n = 4, k = 2;
 
    document.write(FindXorSum(arr, k, n));
 
</script>
Producción: 

11

 

Complejidad de tiempo : O(N)

Espacio Auxiliar: O(N)
 

Publicación traducida automáticamente

Artículo escrito por andrew1234 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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