Given two arrays arr1[] of size M and arr2[] of size N, the task is to find the sum of bitwise OR of each element of arr1[] with every element of the array arr2[].
Examples:
Input: arr1[] = {1, 2, 3}, arr2[] = {1, 2, 3}, M = 3, N = 3
Output: 7 8 9
Explanation:
For arr[0]: Sum = arr1[0]|arr2[0] + arr1[0]|arr2[1] + arr1[0]|arr2[2], Sum = 1|1 + 1|2 + 1|3 = 7
For arr[1], Sum = arr1[1]|arr2[0] + arr1[1]|arr2[1] + arr1[1]|arr2[2], Sum= 2|1 + 2|2 + 2|3 = 8
For arr[2], Sum = arr1[2]|arr2[0] + arr1[2]|arr2[1] + arr1[2]|arr2[2], Sum = 3|1 + 3|2 + 3|3 = 9Input: arr1[] = {2, 4, 8, 16}, arr2[] = {2, 4, 8, 16}, M = 4, N = 4
Output: 36 42 54 78
Naive Approach: The simplest0 approach to solve this problem to traverse the array arr1[] and for each array element in the array arr[], calculate Bitwise OR of each element in the array arr2[].
Time Complexity: O(N2)
Auxiliary Space: O(N)
Efficient Approach: To optimize the above approach, the idea is to use Bit Manipulation to solve the above problem.
- According to the Bitwise OR property, while performing the operation, the ithbit will be set bit only when either of both numbers has a set bit at the ithposition, where 0 ≤ i <32.
- Therefore, for a number in arr1[], if the ith bit is not a set bit, then the ith place will contribute a sum of K * 2i, where K is the total number in arr2[] having set bit at the ith position.
- Otherwise, if the number has a set bit at the ith place, then it will contribute a sum of N * 2i.
Follow the steps below to solve the problem:
- Initialize an integer array, say frequency[], to store the count of numbers in arr2[] having set-bit at ithposition ( 0 ≤ i < 32).
- Traverse the array arr2[] and represent each array element in its binary form and increment the count in the frequency[] array by one at the positions having set bit in the binary representations.
-
Traverse the array arr1[].
- Initialize an integer variable, say bitwise_OR_sum with 0.
- Traverse in the range [0, 31] using variable j.
- If the jth bit is set in the binary representation of arr2[i], then increment bitwise_OR_sum by N * 2j. Otherwise, increment by frequency[j] * 2j
- Print the sum obtained bitwise_OR_sum.
Below is the implementation of the above approach:
// C++ program for the above approach #include <bits/stdc++.h> using namespace std;
// Function to compute sum of Bitwise OR // of each element in arr1[] with all // elements of the array arr2[] void Bitwise_OR_sum_i( int arr1[], int arr2[],
int M, int N)
{ // Declaring an array of
// size 32 to store the
// count of each bit
int frequency[32] = { 0 };
// Traverse the array arr1[]
for ( int i = 0; i < N; i++) {
// Current bit position
int bit_position = 0;
int num = arr1[i];
// While num exceeds 0
while (num) {
// Checks if i-th bit
// is set or not
if (num & 1) {
// Increment the count at
// bit_position by one
frequency[bit_position] += 1;
}
// Increment bit_position
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_OR_sum = 0;
// Traverse in the range [0, 31]
for ( int bit_position = 0;
bit_position < 32;
bit_position++) {
// Check if current bit is set
if (num & 1) {
// Increment the Bitwise
// sum by N*(2^i)
bitwise_OR_sum
+= N * value_at_that_bit;
}
else {
bitwise_OR_sum
+= frequency[bit_position]
* value_at_that_bit;
}
// Right shift num by one
num >>= 1;
// Left shift valee_at_that_bit by one
value_at_that_bit <<= 1;
}
// Print the sum obtained for ith
// number in arr1[]
cout << bitwise_OR_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_OR_sum_i(arr1, arr2, M, N);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to compute sum of Bitwise OR // of each element in arr1[] with all // elements of the array arr2[] static void Bitwise_OR_sum_i( int arr1[], int arr2[],
int M, int N)
{ // Declaring an array of
// size 32 to store the
// count of each bit
int frequency[] = new int [ 32 ];
Arrays.fill(frequency, 0 );
// Traverse the array arr1[]
for ( int i = 0 ; i < N; i++)
{
// Current bit position
int bit_position = 0 ;
int num = arr1[i];
// While num exceeds 0
while (num != 0 )
{
// Checks if i-th bit
// is set or not
if ((num & 1 ) != 0 )
{
// Increment the count at
// bit_position by one
frequency[bit_position] += 1 ;
}
// Increment bit_position
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_OR_sum = 0 ;
// Traverse in the range [0, 31]
for ( int bit_position = 0 ;
bit_position < 32 ;
bit_position++)
{
// Check if current bit is set
if ((num & 1 ) != 0 )
{
// Increment the Bitwise
// sum by N*(2^i)
bitwise_OR_sum += N * value_at_that_bit;
}
else
{
bitwise_OR_sum += frequency[bit_position] *
value_at_that_bit;
}
// Right shift num by one
num >>= 1 ;
// Left shift valee_at_that_bit by one
value_at_that_bit <<= 1 ;
}
// Print the sum obtained for ith
// number in arr1[]
System.out.print(bitwise_OR_sum + " " );
}
return ;
} // 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_OR_sum_i(arr1, arr2, M, N);
} } // This code is contributed by susmitakundugoaldanga |
# Python3 program for the above approach # Function to compute sum of Bitwise OR # of each element in arr1[] with all # elements of the array arr2[] def Bitwise_OR_sum_i(arr1, arr2, M, N):
# Declaring an array of
# size 32 to store the
# count of each bit
frequency = [ 0 ] * 32
# Traverse the array arr1[]
for i in range (N):
# Current bit position
bit_position = 0
num = arr1[i]
# While num exceeds 0
while (num):
# Checks if i-th bit
# is set or not
if (num & 1 ! = 0 ):
# Increment the count at
# bit_position by one
frequency[bit_position] + = 1
# Increment bit_position
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_OR_sum = 0
# Traverse in the range [0, 31]
for bit_position in range ( 32 ):
# Check if current bit is set
if (num & 1 ! = 0 ):
# Increment the Bitwise
# sum by N*(2^i)
bitwise_OR_sum + = N * value_at_that_bit
else :
bitwise_OR_sum + = (frequency[bit_position] *
value_at_that_bit)
# Right shift num by one
num >> = 1
# Left shift valee_at_that_bit by one
value_at_that_bit << = 1
# Print the sum obtained for ith
# number in arr1[]
print (bitwise_OR_sum, end = " " )
return
# Driver Code # 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_OR_sum_i(arr1, arr2, M, N) # This code is contributed by code_hunt |
// C# program for the above approach using System;
class GFG
{ // Function to compute sum of Bitwise OR // of each element in arr1[] with all // elements of the array arr2[] static void Bitwise_OR_sum_i( int [] arr1, int [] arr2,
int M, int N)
{ // Declaring an array of
// size 32 to store the
// count of each bit
int [] frequency = new int [32];
for ( int i = 0; i < 32; i++)
{
frequency[i] = 0;
}
// Traverse the array arr1[]
for ( int i = 0; i < N; i++)
{
// Current bit position
int bit_position = 0;
int num = arr1[i];
// While num exceeds 0
while (num != 0)
{
// Checks if i-th bit
// is set or not
if ((num & 1) != 0)
{
// Increment the count at
// bit_position by one
frequency[bit_position] += 1;
}
// Increment bit_position
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_OR_sum = 0;
// Traverse in the range [0, 31]
for ( int bit_position = 0;
bit_position < 32;
bit_position++)
{
// Check if current bit is set
if ((num & 1) != 0)
{
// Increment the Bitwise
// sum by N*(2^i)
bitwise_OR_sum += N * value_at_that_bit;
}
else
{
bitwise_OR_sum += frequency[bit_position] *
value_at_that_bit;
}
// Right shift num by one
num >>= 1;
// Left shift valee_at_that_bit by one
value_at_that_bit <<= 1;
}
// Print the sum obtained for ith
// number in arr1[]
Console.Write(bitwise_OR_sum + " " );
}
return ;
} // Driver Code public static 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_OR_sum_i(arr1, arr2, M, N);
} } // This code is contributed by sanjoy_62 |
<script> // Javascript program for the above approach // Function to compute sum of Bitwise OR // of each element in arr1[] with all // elements of the array arr2[] function Bitwise_OR_sum_i(arr1, arr2, M, N) {
// Declaring an array of
// size 32 to store the
// count of each bit
let frequency = new Array(32).fill(0);
// Traverse the array arr1[]
for (let i = 0; i < N; i++) {
// Current bit position
let bit_position = 0;
let num = arr1[i];
// While num exceeds 0
while (num) {
// Checks if i-th bit
// is set or not
if (num & 1) {
// Increment the count at
// bit_position by one
frequency[bit_position] += 1;
}
// Increment bit_position
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_OR_sum = 0;
// Traverse in the range [0, 31]
for (let bit_position = 0; bit_position < 32; bit_position++) {
// Check if current bit is set
if (num & 1) {
// Increment the Bitwise
// sum by N*(2^i)
bitwise_OR_sum += N * value_at_that_bit;
}
else {
bitwise_OR_sum += frequency[bit_position] * value_at_that_bit;
}
// Right shift num by one
num >>= 1;
// Left shift valee_at_that_bit by one
value_at_that_bit <<= 1;
}
// Print the sum obtained for ith
// number in arr1[]
document.write(bitwise_OR_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_OR_sum_i(arr1, arr2, M, N); // This code is contributed by _saurabh_jaiswal </script> |
7 8 9
Time Complexity: O(N*32)
Auxiliary Space: O(1) because size of frequency array is constant