Find a number which give minimum sum when XOR with every number of array of integers

Given an array arr[] of non-negative integers, the task is to find an integer X such that (arr[0] XOR X) + (arr[1] XOR X) + … + arr[n – 1] XOR X is minimum possible.
Examples:

Input: arr[] = {3, 9, 6, 2, 4} 
Output: X = 2, Sum = 22
Input: arr[] = {6, 56, 78, 34} 
Output: X = 2, Sum = 170 

Approach: We will check ‘i’th bit of every number of the array in binary representation and count those numbers containing that ‘i’th bit set to ‘1’ because these set bits will contribute to maximize the sum rather than minimize. So we have to make this set ‘i’th bit to ‘0’ if the count is greater than N/2 and if the count is less than N/2 then the numbers having ‘i’th bit set are less and so it will not affect the answer. According to XOR operation on two bits, we know that when A XOR B and both A and B are the same then it gives the result as ‘0’ so we will make that ‘i’th bit in our number (num) to ‘1’, so that (1 XOR 1) will give ‘0’ and minimize the sum.
Below is the implementation of the above approach: 

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
#include <cmath>
using namespace std;

// Function to find an integer X such that
// the sum of all the array elements after
// getting XORed with X is minimum
void findX(int arr[], int n)
{
    // Finding Maximum element of array
    int* itr = max_element(arr, arr + n);

    // Find Maximum number of bits required
    // in the binary representation
    // of maximum number
    // so log2 is calculated
    int p = log2(*itr) + 1;

    // Running loop from p times which is
    // the number of bits required to represent
    // all the elements of the array
    int X = 0;
    for (int i = 0; i < p; i++) {
        int count = 0;
        for (int j = 0; j < n; j++) {

            // If the bits in same position are set
            // then count
            if (arr[j] & (1 << i)) {
                count++;
            }
        }

        // If count becomes greater than half of
        // size of array then we need to make
        // that bit '0' by setting X bit to '1'
        if (count > (n / 2)) {

            // Again using shift operation to calculate
            // the required number
            X += 1 << i;
        }
    }

    // Calculate minimized sum
    long long int sum = 0;
    for (int i = 0; i < n; i++)
        sum += (X ^ arr[i]);

    // Print solution
    cout << "X = " << X << ", Sum = " << sum;
}

// Driver code
int main()
{
    int arr[] = { 2, 3, 4, 5, 6 };
    int n = sizeof(arr) / sizeof(arr[0]);

    findX(arr, n);

    return 0;
}

Java

// Java implementation of above approach
import java.lang.Math;
import java.util.*;

class GFG
{
    // Function to find an integer X such that
    // the sum of all the array elements after
    // getting XORed with X is minimum
    public static void findX(int[] a, int n)
    {

        // Finding Maximum element of array
        Collections.sort(Arrays.asList(a), null);
        int itr = a[n-1];

        // Find Maximum number of bits required
        // in the binary representation
        // of maximum number
        // so log2 is calculated
        int p = (int)(Math.log(itr)/Math.log(2)) + 1;

        // Running loop from p times which is
        // the number of bits required to represent
        // all the elements of the array
        int x = 0;
        for (int i = 0; i < p; i++)
        {
            int count = 0;
            for (int j = 0; j < n; j++)
            {

                // If the bits in same position are set
                // then count
                if ((a[j] & (1 << i)) != 0)
                    count++;
            }

            // If count becomes greater than half of
            // size of array then we need to make
            // that bit '0' by setting X bit to '1'
            if (count > (n / 2))
            {

                // Again using shift operation to calculate
                // the required number
                x += 1 << i;
            }
        }

        // Calculate minimized sum
        long sum = 0;
        for (int i = 0; i < n; i++)
            sum += (x ^ a[i]);

        // Print solution
        System.out.println("X = " + x + ", Sum = " + sum);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int[] a = {2, 3, 4, 5, 6};
        int n = a.length;

        findX(a, n);
    }

}

// This code is contributed by
// sanjeev2552

Python3

# Python 3 implementation of the approach
from math import log2

# Function to find an integer X such that
# the sum of all the array elements after
# getting XORed with X is minimum
def findX(arr, n):

    # Finding Maximum element of array
    itr = arr[0]
    for i in range(len(arr)):

        # Find Maximum number of bits required
        # in the binary representation
        # of maximum number
        # so log2 is calculated
        if(arr[i] > itr):
            itr = arr[i]

    p = int(log2(itr)) + 1

    # Running loop from p times which is
    # the number of bits required to represent
    # all the elements of the array
    X = 0
    for i in range(p):
        count = 0
        for j in range(n):

            # If the bits in same position are set
            # then increase count
            if (arr[j] & (1 << i)):
                count += 1

        # If count becomes greater than half of
        # size of array then we need to make
        # that bit '0' by setting X bit to '1'
        if (count > int(n / 2)):

            # Again using shift operation to calculate
            # the required number
            X += 1 << i

    # Calculate minimized sum
    sum = 0
    for i in range(n):
        sum += (X ^ arr[i])

    # Print solution
    print("X =", X, ", Sum =", sum)

# Driver code
if __name__=='__main__':
    arr = [2, 3, 4, 5, 6]
    n = len(arr)
    findX(arr, n)

# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of above approach
using System;
using System.Linq;

class GFG
{
    // Function to find an integer X such that
    // the sum of all the array elements after
    // getting XORed with X is minimum
    public static void findX(int[] a, int n)
    {

        // Finding Maximum element of array
        int itr = a.Max();

        // Find Maximum number of bits required
        // in the binary representation
        // of maximum number
        // so log2 is calculated
        int p = (int) Math.Log(itr, 2) + 1;

        // Running loop from p times which is
        // the number of bits required to represent
        // all the elements of the array
        int x = 0;
        for (int i = 0; i < p; i++)
        {
            int count = 0;
            for (int j = 0; j < n; j++)
            {

                // If the bits in same position are set
                // then count
                if ((a[j] & (1 << i)) != 0)
                    count++;
            }

            // If count becomes greater than half of
            // size of array then we need to make
            // that bit '0' by setting X bit to '1'
            if (count > (n / 2))
            {

                // Again using shift operation to calculate
                // the required number
                x += 1 << i;
            }
        }

        // Calculate minimized sum
        long sum = 0;
        for (int i = 0; i < n; i++)
            sum += (x ^ a[i]);

        // Print solution
        Console.Write("X = " + x + ", Sum = " + sum);
    }

    // Driver Code
    public static void Main(String[] args)
    {
        int[] a = {2, 3, 4, 5, 6};
        int n = a.Length;

        findX(a, n);
    }

}

// This code is contributed by ravikishor

Javascript

<script>

// Javascript implementation of the approach

// Function to find an integer X such that
// the sum of all the array elements after
// getting XORed with X is minimum
function findX(arr, n)
{
    // Finding Maximum element of array
    let itr = Math.max(...arr);

    // Find Maximum number of bits required
    // in the binary representation
    // of maximum number
    // so log2 is calculated
    let p = parseInt(Math.log(itr) / Math.log(2)) + 1;

    // Running loop from p times which is
    // the number of bits required to represent
    // all the elements of the array
    let X = 0;
    for (let i = 0; i < p; i++) {
        let count = 0;
        for (let j = 0; j < n; j++) {

            // If the bits in same position are set
            // then count
            if (arr[j] & (1 << i)) {
                count++;
            }
        }

        // If count becomes greater than half of
        // size of array then we need to make
        // that bit '0' by setting X bit to '1'
        if (count > parseInt(n / 2)) {

            // Again using shift
            // operation to calculate
            // the required number
            X += 1 << i;
        }
    }

    // Calculate minimized sum
    let sum = 0;
    for (let i = 0; i < n; i++)
        sum += (X ^ arr[i]);

    // Print solution
    document.write("X = " + X + ", Sum = " + sum);
}

// Driver code

    let arr = [ 2, 3, 4, 5, 6 ];
    let n = arr.length;

    findX(arr, n);

</script>

Output:

X = 6, Sum = 14

Time Complexity: O(N * log(A))
Auxiliary Space: O(1)