Dadas dos arrays arr1[] y arr2[] que constan de N y M enteros respectivamente, la tarea es imprimir el Bitwise XOR de Bitwise AND de todos los pares posibles seleccionando un elemento de arr1[] y arr2[].
Ejemplos:
Entrada: arr1[] = {1, 2, 3}, arr2[] = {6, 5}
Salida: 0
Explicación:
AND bit a bit del par (arr1[0], arr2[]) = 1 y 6 = 0.
AND bit a bit del par (arr1[0], arr2[1]) = 1 & 5 = 1.
AND bit a bit del par (arr1[1], arr2[0]) = 2 & 6 = 2.
AND bit a bit del par par (arr1[1], arr2[1]) = 2 y 5 = 0.
AND bit a bit del par (arr1[2], arr2[0]) = 3 & 6 = 2.
AND bit a bit del par (arr1[ 2], arr2[1]) = 3 & 5 = 1.
Por lo tanto, el Bitwise XOR de los valores Bitwise AND obtenidos = 0 ^ 1 ^ 2 ^ 0^ 2 ^ 1 = 0.Entrada: arr1[] = {12}, arr2[] = {4}
Salida: 4
Enfoque ingenuo: el enfoque más simple es encontrar el AND bit a bit de todos los pares posibles seleccionando un elemento de arr1[] y otro elemento de arr2[] y luego, calculando el XOR bit a bit de todos los AND bit a bit de los pares resultantes.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the Bitwise XOR
// of Bitwise AND of all pairs from
// the arrays arr1[] and arr2[]
int findXORS(int arr1[], int arr2[], int N, int M)
{
// Stores the result
int res = 0;
// Iterate over the range [0, N - 1]
for (int i = 0; i < N; i++) {
// Iterate over the range [0, M - 1]
for (int j = 0; j < M; j++) {
// Stores Bitwise AND of
// the pair {arr1[i], arr2[j]}
int temp = arr1[i] & arr2[j];
// Update res
res ^= temp;
}
}
// Return the res
return res;
}
// Driver Code
int main()
{
// Input
int arr1[] = { 1, 2, 3 };
int arr2[] = { 6, 5 };
int N = sizeof(arr1) / sizeof(arr1[0]);
int M = sizeof(arr2) / sizeof(arr2[0]);
cout << findXORS(arr1, arr2, N, M);
return 0;
}
Java
// Java program for the above approach
import java.util.*;
class GFG{
// Function to find the Bitwise XOR
// of Bitwise AND of all pairs from
// the arrays arr1[] and arr2[]
static int findXORS(int arr1[], int arr2[],
int N, int M)
{
// Stores the result
int res = 0;
// Iterate over the range [0, N - 1]
for(int i = 0; i < N; i++)
{
// Iterate over the range [0, M - 1]
for(int j = 0; j < M; j++)
{
// Stores Bitwise AND of
// the pair {arr1[i], arr2[j]}
int temp = arr1[i] & arr2[j];
// Update res
res ^= temp;
}
}
// Return the res
return res;
}
// Driver Code
public static void main(String[] args)
{
// Input
int arr1[] = { 1, 2, 3 };
int arr2[] = { 6, 5 };
int N = arr1.length;
int M = arr2.length;
System.out.print(findXORS(arr1, arr2, N, M));
}
}
// This code is contributed by 29AjayKumar
Python3
# Python 3 program for the above approach
# Function to find the Bitwise XOR
# of Bitwise AND of all pairs from
# the arrays arr1[] and arr2[]
def findXORS(arr1, arr2, N, M):
# Stores the result
res = 0
# Iterate over the range [0, N - 1]
for i in range(N):
# Iterate over the range [0, M - 1]
for j in range(M):
# Stores Bitwise AND of
# the pair {arr1[i], arr2[j]}
temp = arr1[i] & arr2[j]
# Update res
res ^= temp
# Return the res
return res
# Driver Code
if __name__ == '__main__':
# Input
arr1 = [1, 2, 3]
arr2 = [6, 5]
N = len(arr1)
M = len(arr2)
print(findXORS(arr1, arr2, N, M))
# This code is contributed by ipg2016107.
C#
// C# program for the above approach
using System;
class GFG
{
// Function to find the Bitwise XOR
// of Bitwise AND of all pairs from
// the arrays arr1[] and arr2[]
static int findXORS(int[] arr1, int[] arr2, int N,
int M)
{
// Stores the result
int res = 0;
// Iterate over the range [0, N - 1]
for (int i = 0; i < N; i++) {
// Iterate over the range [0, M - 1]
for (int j = 0; j < M; j++)
{
// Stores Bitwise AND of
// the pair {arr1[i], arr2[j]}
int temp = arr1[i] & arr2[j];
// Update res
res ^= temp;
}
}
// Return the res
return res;
}
// Driver Code
public static void Main()
{
// Input
int[] arr1 = { 1, 2, 3 };
int[] arr2 = { 6, 5 };
int N = arr1.Length;
int M = arr2.Length;
Console.Write(findXORS(arr1, arr2, N, M));
}
}
// This code is contributed by ukasp.
Javascript
<script>
// Javascript program for the above approach
// Function to find the Bitwise XOR
// of Bitwise AND of all pairs from
// the arrays arr1[] and arr2[]
function findXORS(arr1, arr2, N, M) {
// Stores the result
let res = 0;
// Iterate over the range [0, N - 1]
for (let i = 0; i < N; i++) {
// Iterate over the range [0, M - 1]
for (let j = 0; j < M; j++) {
// Stores Bitwise AND of
// the pair {arr1[i], arr2[j]}
let temp = arr1[i] & arr2[j];
// Update res
res ^= temp;
}
}
// Return the res
return res;
}
// Driver Code
// Input
let arr1 = [1, 2, 3];
let arr2 = [6, 5];
let N = arr1.length;
let M = arr2.length;
document.write(findXORS(arr1, arr2, N, M));
// This code is contributed by _saurabh_jaiswal
</script>
Producción:
0
Complejidad de Tiempo: O(N * M)
Espacio Auxiliar: O(1)
Enfoque eficiente: el enfoque anterior se puede optimizar en función de las siguientes observaciones:
- Las operaciones Bitwise Xor y Bitwise And tienen propiedades aditivas y distributivas.
- Por lo tanto, considerando las arrays como arr1[] = {A, B} y arr2[] = {X, Y}:
- (A Y X) XOR (A Y Y) XOR (B Y X) XOR (B Y Y)
- (A Y (X XOR Y)) XOR (B Y (X XOR Y))
- (A X O B) Y (X X O Y)
- Por lo tanto, a partir de los pasos anteriores, la tarea se reduce a encontrar el Y bit a bit del XOR bit a bit de arr1[] y arr2[].
Siga los pasos a continuación para resolver el problema:
- Encuentre el Xor bit a bit de cada elemento de una array arr1[] y guárdelo en una variable, digamos XORS1.
- Encuentre el Xor bit a bit de cada elemento de una array arr2[] y guárdelo en una variable, digamos XORS2 .
- Finalmente, imprima el resultado como AND bit a bit de XORS1 y XORS2.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to find the Bitwise XOR
// of Bitwise AND of all pairs from
// the arrays arr1[] and arr2[]
int findXORS(int arr1[], int arr2[],
int N, int M)
{
// Stores XOR of array arr1[]
int XORS1 = 0;
// Stores XOR of array arr2[]
int XORS2 = 0;
// Traverse the array arr1[]
for (int i = 0; i < N; i++) {
XORS1 ^= arr1[i];
}
// Traverse the array arr2[]
for (int i = 0; i < M; i++) {
XORS2 ^= arr2[i];
}
// Return the result
return XORS1 and XORS2;
}
// Driver Code
int main()
{
// Input
int arr1[] = { 1, 2, 3 };
int arr2[] = { 6, 5 };
int N = sizeof(arr1) / sizeof(arr1[0]);
int M = sizeof(arr2) / sizeof(arr2[0]);
cout << findXORS(arr1, arr2, N, M);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
class GFG{
// Function to find the Bitwise XOR
// of Bitwise AND of all pairs from
// the arrays arr1[] and arr2[]
static int findXORS(int arr1[], int arr2[],
int N, int M)
{
// Stores XOR of array arr1[]
int XORS1 = 0;
// Stores XOR of array arr2[]
int XORS2 = 0;
// Traverse the array arr1[]
for(int i = 0; i < N; i++)
{
XORS1 ^= arr1[i];
}
// Traverse the array arr2[]
for(int i = 0; i < M; i++)
{
XORS2 ^= arr2[i];
}
// Return the result
return (XORS1 & XORS2);
}
// Driver Code
public static void main(String[] args)
{
// Input
int arr1[] = { 1, 2, 3 };
int arr2[] = { 6, 5 };
int N = arr1.length;
int M = arr2.length;
System.out.println(findXORS(arr1, arr2, N, M));
}
}
// This code is contributed by susmitakundugoaldanga
Python3
# Python3 program for the above approach # Function to find the Bitwise XOR # of Bitwise AND of all pairs from # the arrays arr1[] and arr2[] def findXORS(arr1, arr2, N, M): # Stores XOR of array arr1[] XORS1 = 0 # Stores XOR of array arr2[] XORS2 = 0 # Traverse the array arr1[] for i in range(N): XORS1 ^= arr1[i] # Traverse the array arr2[] for i in range(M): XORS2 ^= arr2[i] # Return the result return XORS1 and XORS2 # Driver Code if __name__ == '__main__': # Input arr1 = [ 1, 2, 3 ] arr2 = [ 6, 5 ] N = len(arr1) M = len(arr2) print(findXORS(arr1, arr2, N, M)) # This code is contributed by bgangwar59
C#
// C# program for the above approach
using System;
class GFG{
// Function to find the Bitwise XOR
// of Bitwise AND of all pairs from
// the arrays arr1[] and arr2[]
static int findXORS(int []arr1, int []arr2,
int N, int M)
{
// Stores XOR of array arr1[]
int XORS1 = 0;
// Stores XOR of array arr2[]
int XORS2 = 0;
// Traverse the array arr1[]
for(int i = 0; i < N; i++)
{
XORS1 ^= arr1[i];
}
// Traverse the array arr2[]
for(int i = 0; i < M; i++)
{
XORS2 ^= arr2[i];
}
// Return the result
return (XORS1 & XORS2);
}
// Driver Code
public static void Main(String[] args)
{
// Input
int []arr1 = { 1, 2, 3 };
int []arr2 = { 6, 5 };
int N = arr1.Length;
int M = arr2.Length;
Console.WriteLine(findXORS(arr1, arr2, N, M));
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript program for the above approach
// Function to find the Bitwise XOR
// of Bitwise AND of all pairs from
// the arrays arr1[] and arr2[]
function findXORS(arr1, arr2, N, M)
{
// Stores XOR of array arr1[]
let XORS1 = 0;
// Stores XOR of array arr2[]
let XORS2 = 0;
// Traverse the array arr1[]
for (let i = 0; i < N; i++) {
XORS1 ^= arr1[i];
}
// Traverse the array arr2[]
for (let i = 0; i < M; i++) {
XORS2 ^= arr2[i];
}
// Return the result
return XORS1 && XORS2;
}
// Driver Code
// Input
let arr1 = [ 1, 2, 3 ];
let arr2 = [ 6, 5 ];
let N = arr1.length;
let M = arr2.length;
document.write(findXORS(arr1, arr2, N, M));
// This code is contributed by Dharanendra L V.
</script>
Producción:
0
Complejidad temporal: O(N + M)
Espacio auxiliar: O(1)
Publicación traducida automáticamente
Artículo escrito por srinivasteja18 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA