Dadas dos arrays arr1[] de tamaño M y arr2[] de tamaño N , la tarea es encontrar la suma de AND bit a bit de cada elemento de arr1[] con los elementos de la array arr2[] .
Ejemplos:
Entrada: arr1[] = {1, 2, 3}, arr2[] = {1, 2, 3}, M = 3, N = 3
Salida: 2 4 6
Explicación:
Para elementos en el índice 0 en arr1[], Suma = arr1[0] & arr2[0] + arr1[0] & arr2[1] + arr1[0] & arr2[2], Suma = 1 & 1 + 1 & 2 + 1 & 3 = 2
Para elementos en índice 1 en arr1[], Sum = arr1[1] & arr2[0] + arr1[1] & arr2[1] + arr1[1] & arr2[2], Sum= 2 & 1 + 2 & 2 + 2 & 3 = 4
Para elementos en el índice 2 en arr1[], Sum = arr1[2] & arr2[0] + arr1[2] & arr2[1] + arr1[2] & arr2[2], Sum= 3 & 1 + 3 y 2 + 3 y 3 = 6Entrada: arr1[] = {2, 4, 8, 16}, arr2[] = {2, 4, 8, 16}, M = 4, N = 4
Salida: 2 4 8 16
Enfoque ingenuo: el enfoque más simple para resolver el problema es recorrer la array arr1[] y, para cada elemento de la array arr1[] , recorrer la array arr2[] y calcular la suma de Bitwise Y de los elementos actuales de arr1[] con todos los elementos de arr2[] para cada elemento
Tiempo Complejidad: O(N 2 )
Espacio Auxiliar: O(N)
Enfoque eficiente: la idea es utilizar la manipulación de bits para resolver el problema anterior. Suponga que cada elemento de la array se puede representar usando solo 32 bits.
- De acuerdo con la propiedad AND bit a bit , mientras se realiza la operación, el i -ésimo bit se establecerá bit solo cuando ambos números tengan un bit establecido en la i -ésima posición, donde 0≤i<32 .
- Por lo tanto, para un número en arr1[], si el i -ésimo bit es un bit establecido, entonces el i -ésimo lugar contribuirá con una suma de K*2 i , donde K es el número total de números en arr2[] habiendo establecido el bit en la i -ésima posición.
Siga los pasos a continuación para resolver el problema:
- Inicialice una frecuencia de array de enteros [] para almacenar el recuento de números en arr2 [] habiendo establecido el bit en la i -ésima posición donde 0≤i<32
- Recorra la array arr2[] y para cada elemento represéntelo en forma binaria e incremente el conteo en la array de frecuencia[] en uno en la posición que tiene 1 en la representación binaria .
- Atraviesa la array arr1[]
- Inicialice una variable entera bitwise_AND_sum con 0 .
- Traverse en el rango [0, 31] usando una variable j .
- Si el j -ésimo bit se establece en bit en la representación binaria de arr2[i] , entonces incremente bitwise_AND_sum por la frecuencia[j]*2 j .
- Imprime la suma obtenida, es decir, bitwise_AND_sum .
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 compute the AND sum
// for each element of an array
void Bitwise_AND_sum_i(int arr1[], int arr2[], int M, int N)
{
// Declaring an array of
// size 32 for storing the
// count of each bit
int frequency[32] = { 0 };
// Traverse the array arr2[]
// and store the count of a
// bit in frequency array
for (int i = 0; i < N; i++) {
// Current bit position
int bit_position = 0;
int num = arr1[i];
// While num is greater
// than 0
while (num) {
// Checks if ith bit is
// set or not
if (num & 1) {
// Increment the count of
// bit by one
frequency[bit_position] += 1;
}
// Increment the bit position
// by one
bit_position += 1;
// Right shift the num by one
num >>= 1;
}
}
// Traverse in the arr2[]
for (int i = 0; i < M; i++) {
int num = arr2[i];
// Store the ith bit
// value
int value_at_that_bit = 1;
// Total required sum
int bitwise_AND_sum = 0;
// Traverse in the range [0, 31]
for (int bit_position = 0; bit_position < 32;
bit_position++) {
// Checks if current bit is set
if (num & 1) {
// Increment the bitwise sum
// by frequency[bit_position]
// * value_at_that_bit;
bitwise_AND_sum += frequency[bit_position]
* value_at_that_bit;
}
// Right shift num by one
num >>= 1;
// Left shift vale_at_that_bit by one
value_at_that_bit <<= 1;
}
// Print the sum obtained for ith
// number in arr1[]
cout << bitwise_AND_sum << ' ';
}
return;
}
// Driver Code
int main()
{
// Given arr1[]
int arr1[] = { 1, 2, 3 };
// Given arr2[]
int arr2[] = { 1, 2, 3 };
// Size of arr1[]
int N = sizeof(arr1) / sizeof(arr1[0]);
// Size of arr2[]
int M = sizeof(arr2) / sizeof(arr2[0]);
// Function Call
Bitwise_AND_sum_i(arr1, arr2, M, N);
return 0;
}
Java
// Java program for the above approach
import java.io.*;
class GFG
{
// Driver Code
public static void main(String[] args)
{
// Given arr1[]
int[] arr1 = { 1, 2, 3 };
// Given arr2[]
int[] arr2 = { 1, 2, 3 };
// Size of arr1[]
int N = arr1.length;
// Size of arr2[]
int M = arr2.length;
// Function Call
Bitwise_AND_sum_i(arr1, arr2, M, N);
}
// Function to compute the AND sum
// for each element of an array
static void Bitwise_AND_sum_i(int arr1[], int arr2[],
int M, int N)
{
// Declaring an array of
// size 32 for storing the
// count of each bit
int[] frequency = new int[32];
// Traverse the array arr2[]
// and store the count of a
// bit in frequency array
for (int i = 0; i < N; i++)
{
// Current bit position
int bit_position = 0;
int num = arr1[i];
// While num is greater
// than 0
while (num != 0)
{
// Checks if ith bit is
// set or not
if ((num & 1) != 0)
{
// Increment the count of
// bit by one
frequency[bit_position] += 1;
}
// Increment the bit position
// by one
bit_position += 1;
// Right shift the num by one
num >>= 1;
}
}
// Traverse in the arr2[]
for (int i = 0; i < M; i++)
{
int num = arr2[i];
// Store the ith bit
// value
int value_at_that_bit = 1;
// Total required sum
int bitwise_AND_sum = 0;
// Traverse in the range [0, 31]
for (int bit_position = 0; bit_position < 32;
bit_position++)
{
// Checks if current bit is set
if ((num & 1) != 0)
{
// Increment the bitwise sum
// by frequency[bit_position]
// * value_at_that_bit;
bitwise_AND_sum
+= frequency[bit_position]
* value_at_that_bit;
}
// Right shift num by one
num >>= 1;
// Left shift vale_at_that_bit by one
value_at_that_bit <<= 1;
}
// Print the sum obtained for ith
// number in arr1[]
System.out.print( bitwise_AND_sum + " ");
}
}
}
// This code is contributed by Dharanendra L V
Python3
# Python3 program for the above approach # Function to compute the AND sum # for each element of an array def Bitwise_AND_sum_i(arr1, arr2, M, N): # Declaring an array of # size 32 for storing the # count of each bit frequency = [0]*32 # Traverse the array arr2[] # and store the count of a # bit in frequency array for i in range(N): # Current bit position bit_position = 0 num = arr1[i] # While num is greater # than 0 while (num): # Checks if ith bit is # set or not if (num & 1): # Increment the count of # bit by one frequency[bit_position] += 1 # Increment the bit position # by one bit_position += 1 # Right shift the num by one num >>= 1 # Traverse in the arr2[] for i in range(M): num = arr2[i] # Store the ith bit # value value_at_that_bit = 1 # Total required sum bitwise_AND_sum = 0 # Traverse in the range [0, 31] for bit_position in range(32): # Checks if current bit is set if (num & 1): # Increment the bitwise sum # by frequency[bit_position] # * value_at_that_bit bitwise_AND_sum += frequency[bit_position] * value_at_that_bit # Right shift num by one num >>= 1 # Left shift vale_at_that_bit by one value_at_that_bit <<= 1 # Print sum obtained for ith # number in arr1[] print(bitwise_AND_sum, end = " ") return # Driver Code if __name__ == '__main__': # Given arr1[] arr1 = [1, 2, 3] # Given arr2 arr2 = [1, 2, 3] # Size of arr1[] N = len(arr1) # Size of arr2[] M = len(arr2) # Function Call Bitwise_AND_sum_i(arr1, arr2, M, N) # This code is contributed by mohit kumar 29
C#
// C# program for the above approach
using System;
class GFG
{
// Driver code
static public void Main()
{
// Given arr1[]
int[] arr1 = { 1, 2, 3 };
// Given arr2[]
int[] arr2 = { 1, 2, 3 };
// Size of arr1[]
int N = arr1.Length;
// Size of arr2[]
int M = arr2.Length;
// Function Call
Bitwise_AND_sum_i(arr1, arr2, M, N);
}
// Function to compute the AND sum
// for each element of an array
static void Bitwise_AND_sum_i(int[] arr1, int[] arr2,
int M, int N)
{
// Declaring an array of
// size 32 for storing the
// count of each bit
int[] frequency = new int[32];
// Traverse the array arr2[]
// and store the count of a
// bit in frequency array
for (int i = 0; i < N; i++)
{
// Current bit position
int bit_position = 0;
int num = arr1[i];
// While num is greater
// than 0
while (num != 0)
{
// Checks if ith bit is
// set or not
if ((num & 1) != 0)
{
// Increment the count of
// bit by one
frequency[bit_position] += 1;
}
// Increment the bit position
// by one
bit_position += 1;
// Right shift the num by one
num >>= 1;
}
}
// Traverse in the arr2[]
for (int i = 0; i < M; i++)
{
int num = arr2[i];
// Store the ith bit
// value
int value_at_that_bit = 1;
// Total required sum
int bitwise_AND_sum = 0;
// Traverse in the range [0, 31]
for (int bit_position = 0; bit_position < 32;
bit_position++) {
// Checks if current bit is set
if ((num & 1) != 0)
{
// Increment the bitwise sum
// by frequency[bit_position]
// * value_at_that_bit;
bitwise_AND_sum
+= frequency[bit_position]
* value_at_that_bit;
}
// Right shift num by one
num >>= 1;
// Left shift vale_at_that_bit by one
value_at_that_bit <<= 1;
}
// Print the sum obtained for ith
// number in arr1[]
Console.Write(bitwise_AND_sum + " ");
}
}
}
// The code is contributed by Dharanendra L V
Javascript
<script>
// Javascript program for the above approach
// Function to compute the AND sum
// for each element of an array
function Bitwise_AND_sum_i(arr1, arr2, M, N) {
// Declaring an array of
// size 32 for storing the
// count of each bit
let frequency = new Array(32).fill(0);
// Traverse the array arr2[]
// and store the count of a
// bit in frequency array
for (let i = 0; i < N; i++) {
// Current bit position
let bit_position = 0;
let num = arr1[i];
// While num is greater
// than 0
while (num) {
// Checks if ith bit is
// set or not
if (num & 1) {
// Increment the count of
// bit by one
frequency[bit_position] += 1;
}
// Increment the bit position
// by one
bit_position += 1;
// Right shift the num by one
num >>= 1;
}
}
// Traverse in the arr2[]
for (let i = 0; i < M; i++) {
let num = arr2[i];
// Store the ith bit
// value
let value_at_that_bit = 1;
// Total required sum
let bitwise_AND_sum = 0;
// Traverse in the range [0, 31]
for (let bit_position = 0; bit_position < 32; bit_position++) {
// Checks if current bit is set
if (num & 1) {
// Increment the bitwise sum
// by frequency[bit_position]
// * value_at_that_bit;
bitwise_AND_sum += frequency[bit_position] * value_at_that_bit;
}
// Right shift num by one
num >>= 1;
// Left shift vale_at_that_bit by one
value_at_that_bit <<= 1;
}
// Print the sum obtained for ith
// number in arr1[]
document.write(bitwise_AND_sum + ' ');
}
return;
}
// Driver Code
// Given arr1[]
let arr1 = [1, 2, 3];
// Given arr2[]
let arr2 = [1, 2, 3];
// Size of arr1[]
let N = arr1.length;
// Size of arr2[]
let M = arr2.length
// Function Call
Bitwise_AND_sum_i(arr1, arr2, M, N);
// This code is contributed by _saurabh_jaiswal
</script>
Producción:
2 4 6
Complejidad de Tiempo: O(N * 32)
Espacio Auxiliar: O(N * 32)