Count of even and odd set bit Array elements after XOR with K for Q queries

Given an array arr of N elements and another array Q containing values of K, the task is to print the count of elements in the array arr with odd and even set bits after its XOR with each element K in the array Q.

Examples:

Input: arr[] = { 2, 7, 4, 5, 3 }, Q[] = { 3, 4, 12, 6 } 
Output: 2 3 
3 2 
2 3 
2 3

Input: arr[] = { 7, 1, 6, 5, 11 }, Q[] = { 2, 10, 3, 6 } 
Output: 3 2 
2 3 
2 3 
2 3  

Approach:

• XOR of two elements both having odd or even set bits, results to even set bits.
• XOR of two elements, one having odd and other having even set bits or vice versa, results to odd set bits.
• Precompute count of elements with even and odd set bits of all array elements using Brian Kernighan’s Algorithm.
• For all elements of Q, count number of set bits. If count of set bits is even, the count of even and odd set bits elements remain unchanged. Otherwise reverse the count and display.

Below is the implementation of the above approach: 

C++

// C++ Program to count number
// of even and odd set bits
// elements after XOR with a
// given element

#include <bits/stdc++.h>
using namespace std;

void keep_count(int arr[], int& even,
                int& odd, int N)
{
    // Store the count of set bits
    int count;
    for (int i = 0; i < N; i++) {
        count = 0;

        // Brian Kernighan's algorithm
        while (arr[i] != 0) {
            arr[i] = (arr[i] - 1) & arr[i];
            count++;
        }

        if (count % 2 == 0)
            even++;
        else
            odd++;
    }

    return;
}

// Function to solve Q queries
void solveQueries(
    int arr[], int n,
    int q[], int m)
{

    int even_count = 0, odd_count = 0;

    keep_count(arr, even_count,
               odd_count, n);

    for (int i = 0; i < m; i++) {

        int X = q[i];

        // Store set bits in X
        int count = 0;

        // Count set bits of X
        while (X != 0) {
            X = (X - 1) & X;
            count++;
        }

        if (count % 2 == 0) {
            cout << even_count << " "
                 << odd_count << "\n";
        }
        else {
            cout << odd_count << " "
                 << even_count
                 << "\n";
        }
    }
}

// Driver code
int main()
{
    int arr[] = { 2, 7, 4, 5, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);

    int q[] = { 3, 4, 12, 6 };
    int m = sizeof(q) / sizeof(q[0]);

    solveQueries(arr, n, q, m);

    return 0;
}

Java

// Java program to count number
// of even and odd set bits
// elements after XOR with a
// given element
class GFG{

static int even, odd;

static void keep_count(int arr[], int N)
{

    // Store the count of set bits
    int count;

    for(int i = 0; i < N; i++)
    {
       count = 0;

       // Brian Kernighan's algorithm
       while (arr[i] != 0)
       {
           arr[i] = (arr[i] - 1) & arr[i];
           count++;
       }
       if (count % 2 == 0)
           even++;
       else
           odd++;
    }
    return;
}

// Function to solve Q queries
static void solveQueries(int arr[], int n,
                         int q[], int m)
{
    even = 0;
    odd = 0;
    keep_count(arr, n);

    for(int i = 0; i < m; i++)
    {
       int X = q[i];

       // Store set bits in X
       int count = 0;

       // Count set bits of X
       while (X != 0)
       {
           X = (X - 1) & X;
           count++;
       }
       if (count % 2 == 0)
       {
           System.out.print(even + " " +
                             odd + "\n");
       }
       else
       {
           System.out.print(odd + " " +
                           even + "\n");
       }
    }
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 7, 4, 5, 3 };
    int n = arr.length;

    int q[] = { 3, 4, 12, 6 };
    int m = q.length;

    solveQueries(arr, n, q, m);
}
}

// This code is contributed by amal kumar choubey

Python3

# Python3 program to count number
# of even and odd set bits
# elements after XOR with a
# given element

even = 0
odd = 0

def keep_count(arr, N):

    global even
    global odd

    # Store the count of set bits
    for i in range(N):
        count = 0

        # Brian Kernighan's algorithm
        while (arr[i] != 0):
            arr[i] = (arr[i] - 1) & arr[i]
            count += 1

        if (count % 2 == 0):
            even += 1
        else:
            odd += 1

    return

# Function to solve Q queries
def solveQueries(arr, n, q, m):

    global even
    global odd

    keep_count(arr, n)

    for i in range(m):
        X = q[i]

        # Store set bits in X
        count = 0

        # Count set bits of X
        while (X != 0):
            X = (X - 1) & X
            count += 1

        if (count % 2 == 0):
            print(even, odd)
        else:
            print(odd, even)

# Driver code
if __name__ == '__main__':

    arr = [ 2, 7, 4, 5, 3 ]
    n = len(arr)

    q = [ 3, 4, 12, 6 ]
    m = len(q)

    solveQueries(arr, n, q, m)

# This code is contributed by samarth

C#

// C# program to count number
// of even and odd set bits
// elements after XOR with a
// given element
using System;
class GFG{

static int even, odd;

static void keep_count(int []arr, int N)
{

    // Store the count of set bits
    int count;

    for(int i = 0; i < N; i++)
    {
        count = 0;

        // Brian Kernighan's algorithm
        while (arr[i] != 0)
        {
            arr[i] = (arr[i] - 1) & arr[i];
            count++;
        }
        if (count % 2 == 0)
            even++;
        else
            odd++;
    }
    return;
}

// Function to solve Q queries
static void solveQueries(int []arr, int n,
                         int []q, int m)
{
    even = 0;
    odd = 0;
    keep_count(arr, n);

    for(int i = 0; i < m; i++)
    {
        int X = q[i];

        // Store set bits in X
        int count = 0;

        // Count set bits of X
        while (X != 0)
        {
            X = (X - 1) & X;
            count++;
        }
        if (count % 2 == 0)
        {
            Console.Write(even + " " +
                           odd + "\n");
        }
        else
        {
            Console.Write(odd + " " +
                         even + "\n");
        }
    }
}

// Driver code
public static void Main()
{
    int []arr = { 2, 7, 4, 5, 3 };
    int n = arr.Length;

    int []q = { 3, 4, 12, 6 };
    int m = q.Length;

    solveQueries(arr, n, q, m);
}
}

// This code is contributed by Code_Mech

Javascript

<script>

// Javascript program to count number
// of even and odd set bits
// elements after XOR with a
// given element function even, odd;
function keep_count(arr, N)
{

    // Store the count of set bits
    var count;

    for(i = 0; i < N; i++)
    {
        count = 0;

        // Brian Kernighan's algorithm
        while (arr[i] != 0)
        {
            arr[i] = (arr[i] - 1) & arr[i];
            count++;
        }
        if (count % 2 == 0)
            even++;
        else
            odd++;
    }
    return;
}

// Function to solve Q queries
function solveQueries(arr, n, q, m)
{
    even = 0;
    odd = 0;
    keep_count(arr, n);

    for(i = 0; i < m; i++)
    {
        var X = q[i];

        // Store set bits in X
        var count = 0;

        // Count set bits of X
        while (X != 0)
        {
            X = (X - 1) & X;
            count++;
        }
        if (count % 2 == 0)
        {
            document.write(even + " " +
                            odd + "<br/>");
        }
        else
        {
            document.write(odd + " " +
                          even + "<br/>");
        }
    }
}

// Driver code
var arr = [ 2, 7, 4, 5, 3 ];
var n = arr.length;

var q = [ 3, 4, 12, 6 ];
var m = q.length;

solveQueries(arr, n, q, m);

// This code is contributed by aashish1995

</script>

Output:

2 3
3 2
2 3
2 3

Time Complexity: O(N * log(max(arr))+m*log(max(q))) where max(arr) and max(q) represents the maximum element in array arr and q respectively.
Auxiliary Space: O(1) as constant space is used