Sum of Bitwise XOR of elements of an array with all elements of another array

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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 i^th 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 i^th bit not set * 2ⁱto 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

// 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

# 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#

// 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

Javascript

<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>

Output:

7 34 11

Time Complexity: O(N)
Auxiliary Space: O(N)

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