Sum of Bitwise XOR of elements of an array with all elements of another array
Last Updated :
28 Jun, 2021
Given an array arr[] of size N and an array Q[], the task is to calculate the sum of Bitwise XOR of all elements of the array arr[] with each element of the array q[].
Examples:
Input: arr[ ] = {5, 2, 3}, Q[ ] = {3, 8, 7}
Output: 7 34 11
Explanation:
For Q[0] ( = 3): Sum = 5 ^ 3 + 2 ^ 3 + 3 ^ 3 = 7.
For Q[1] ( = 8): Sum = 5 ^ 8 + 2 ^ 8 + 3 ^ 8 = 34.
For Q[2] ( = 7): Sum = 5 ^ 7 + 2 ^ 7 + 3 ^ 7 = 11.
Input: arr[ ] = {2, 3, 4}, Q[ ] = {1, 2}
Output: 10 7
Naive Approach: The simplest approach to solve the problem is to traverse the array Q[] and for each array element, calculate the sum of its Bitwise XOR with all elements of the array arr[].
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: Follow the steps below to optimize the above approach:
- Initialize an array count[], of size 32. to store the count of set bits at each position of the elements of the array arr[].
- Traverse the array arr[].
- Update the array count[] accordingly. In a 32-bit binary representation, if the ith bit is set, increase the count of set bits at that position.
- Traverse the array Q[] and for each array element, perform the following operations:
- Initialize variables, say sum = 0, to store the required sum of Bitwise XOR .
- Iterate over each bit positions of the current element.
- If current bit is set, add count of elements with ith bit not set * 2i to sum.
- Otherwise, add count[i] * 2i.
- Finally, print the value of sum.
Below is the implementation of the above approach:
C++
// C++ Program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to calculate sum of Bitwise
// XOR of elements of arr[] with k
int xorSumOfArray(int arr[], int n, int k, int count[])
{
// Initialize sum to be zero
int sum = 0;
int p = 1;
// Iterate over each set bit
for (int i = 0; i < 31; i++) {
// Stores contribution of
// i-th bet to the sum
int val = 0;
// If the i-th bit is set
if ((k & (1 << i)) != 0) {
// Stores count of elements
// whose i-th bit is not set
int not_set = n - count[i];
// Update value
val = ((not_set)*p);
}
else {
// Update value
val = (count[i] * p);
}
// Add value to sum
sum += val;
// Move to the next
// power of two
p = (p * 2);
}
return sum;
}
void sumOfXors(int arr[], int n, int queries[], int q)
{
// Stores the count of elements
// whose i-th bit is set
int count[32];
// Initialize count to 0
// for all positions
memset(count, 0, sizeof(count));
// Traverse the array
for (int i = 0; i < n; i++) {
// Iterate over each bit
for (int j = 0; j < 31; j++) {
// If the i-th bit is set
if (arr[i] & (1 << j))
// Increase count
count[j]++;
}
}
for (int i = 0; i < q; i++) {
int k = queries[i];
cout << xorSumOfArray(arr, n, k, count) << " ";
}
}
// Driver Code
int main()
{
int arr[] = { 5, 2, 3 };
int queries[] = { 3, 8, 7 };
int n = sizeof(arr) / sizeof(int);
int q = sizeof(queries) / sizeof(int);
sumOfXors(arr, n, queries, q);
return 0;
}
// Java Program for the above approach
import java.util.Arrays;
class GFG{
// Function to calculate sum of Bitwise
// XOR of elements of arr[] with k
static int xorSumOfArray(int arr[], int n,
int k, int count[])
{
// Initialize sum to be zero
int sum = 0;
int p = 1;
// Iterate over each set bit
for(int i = 0; i < 31; i++)
{
// Stores contribution of
// i-th bet to the sum
int val = 0;
// If the i-th bit is set
if ((k & (1 << i)) != 0)
{
// Stores count of elements
// whose i-th bit is not set
int not_set = n - count[i];
// Update value
val = ((not_set)*p);
}
else
{
// Update value
val = (count[i] * p);
}
// Add value to sum
sum += val;
// Move to the next
// power of two
p = (p * 2);
}
return sum;
}
static void sumOfXors(int arr[], int n,
int queries[], int q)
{
// Stores the count of elements
// whose i-th bit is set
int []count = new int[32];
// Initialize count to 0
// for all positions
Arrays.fill(count,0);
// Traverse the array
for(int i = 0; i < n; i++)
{
// Iterate over each bit
for(int j = 0; j < 31; j++)
{
// If the i-th bit is set
if ((arr[i] & (1 << j)) != 0)
// Increase count
count[j]++;
}
}
for(int i = 0; i < q; i++)
{
int k = queries[i];
System.out.print(
xorSumOfArray(arr, n, k, count) + " ");
}
}
// Driver Code
public static void main(String args[])
{
int arr[] = { 5, 2, 3 };
int queries[] = { 3, 8, 7 };
int n = arr.length;
int q = queries.length;
sumOfXors(arr, n, queries, q);
}
}
// This code is contributed by SoumikMondal
# Python3 Program for the above approach
# Function to calculate sum of Bitwise
# XOR of elements of arr[] with k
def xorSumOfArray(arr, n, k, count):
# Initialize sum to be zero
sum = 0
p = 1
# Iterate over each set bit
for i in range(31):
# Stores contribution of
# i-th bet to the sum
val = 0
# If the i-th bit is set
if ((k & (1 << i)) != 0):
# Stores count of elements
# whose i-th bit is not set
not_set = n - count[i]
# Update value
val = ((not_set)*p)
else:
# Update value
val = (count[i] * p)
# Add value to sum
sum += val
# Move to the next
# power of two
p = (p * 2)
return sum
def sumOfXors(arr, n, queries, q):
# Stores the count of elements
# whose i-th bit is set
count = [0 for i in range(32)]
# Traverse the array
for i in range(n):
# Iterate over each bit
for j in range(31):
# If the i-th bit is set
if (arr[i] & (1 << j)):
# Increase count
count[j] += 1
for i in range(q):
k = queries[i]
print(xorSumOfArray(arr, n, k, count), end = " ")
# Driver Code
if __name__ == '__main__':
arr = [ 5, 2, 3 ]
queries = [ 3, 8, 7 ]
n = len(arr)
q = len(queries)
sumOfXors(arr, n, queries, q)
# This code is contributed by SURENDRA_GANGWAR
// C# Program for the above approach
using System;
public class GFG{
// Function to calculate sum of Bitwise
// XOR of elements of arr[] with k
static int xorSumOfArray(int []arr, int n, int k, int []count)
{
// Initialize sum to be zero
int sum = 0;
int p = 1;
// Iterate over each set bit
for (int i = 0; i < 31; i++) {
// Stores contribution of
// i-th bet to the sum
int val = 0;
// If the i-th bit is set
if ((k & (1 << i)) != 0) {
// Stores count of elements
// whose i-th bit is not set
int not_set = n - count[i];
// Update value
val = ((not_set)*p);
}
else {
// Update value
val = (count[i] * p);
}
// Add value to sum
sum += val;
// Move to the next
// power of two
p = (p * 2);
}
return sum;
}
static void sumOfXors(int []arr, int n, int []queries, int q)
{
// Stores the count of elements
// whose i-th bit is set
int []count = new int[32];
// Initialize count to 0
// for all positions
for(int i = 0; i < 32; i++)
count[i] = 0;
// Traverse the array
for (int i = 0; i < n; i++) {
// Iterate over each bit
for (int j = 0; j < 31; j++) {
// If the i-th bit is set
if ((arr[i] & (1 << j)) != 0)
// Increase count
count[j]++;
}
}
for (int i = 0; i < q; i++) {
int k = queries[i];
Console.Write(xorSumOfArray(arr, n, k, count) + " ");
}
}
// Driver Code
static public void Main ()
{
int []arr = { 5, 2, 3 };
int []queries = { 3, 8, 7 };
int n = arr.Length;
int q = queries.Length;
sumOfXors(arr, n, queries, q);
}
}
// This code is contributed by AnkThon
<script>
// Javascript program for the above approach
// Function to calculate sum of Bitwise
// XOR of elements of arr[] with k
function xorSumOfArray(arr, n, k, count)
{
// Initialize sum to be zero
var sum = 0;
var p = 1;
// Iterate over each set bit
for(var i = 0; i < 31; i++)
{
// Stores contribution of
// i-th bet to the sum
var val = 0;
// If the i-th bit is set
if ((k & (1 << i)) != 0)
{
// Stores count of elements
// whose i-th bit is not set
var not_set = n - count[i];
// Update value
val = ((not_set)*p);
}
else
{
// Update value
val = (count[i] * p);
}
// Add value to sum
sum += val;
// Move to the next
// power of two
p = (p * 2);
}
return sum;
}
function sumOfXors(arr, n, queries, q)
{
// Stores the count of elements
// whose i-th bit is set
var count = new Array(32);
// Initialize count to 0
// for all positions
count.fill(0);
// Traverse the array
for(var i = 0; i < n; i++)
{
// Iterate over each bit
for(var j = 0; j < 31; j++)
{
// If the i-th bit is set
if (arr[i] & (1 << j))
// Increase count
count[j]++;
}
}
for(var i = 0; i < q; i++)
{
var k = queries[i];
document.write(xorSumOfArray(
arr, n, k, count) + " ");
}
}
// Driver code
var arr = [ 5, 2, 3 ];
var queries = [ 3, 8, 7 ];
var n = arr.length;
var q = queries.length;
sumOfXors(arr, n, queries, q);
// This code is contributed by SoumikMondal
</script>
Time Complexity: O(N)
Auxiliary Space: O(N)
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