Find the missing element in an array of integers represented in binary format

// C++ program to find the missing integer
// in N numbers when N bits are given
#include <bits/stdc++.h>
using namespace std;
 
class BitInteger {
private:
    bool* bits;
 
public:
    static const int INTEGER_SIZE = 32;
 
    BitInteger()
    {
        bits = new bool[INTEGER_SIZE];
    }
 
    // Constructor to convert an integer
    // variable into binary format
    BitInteger(int value)
    {
        bits = new bool[INTEGER_SIZE];
 
        for (int j = 0; j < INTEGER_SIZE; j++) {
 
            // The if statement will shift the
            // original value j times.
            // So that appropriate (INTEGER_SIZE - 1 -j)th
            // bits will be either 0/1.
            //  (INTEGER_SIZE - 1 -j)th bit for all
            // j = 0 to INTEGER_SIZE-1 corresponds
            // to  LSB to MSB respectively.
            if (((value >> j) & 1) == 1)
                bits[INTEGER_SIZE - 1 - j] = true;
            else
                bits[INTEGER_SIZE - 1 - j] = false;
        }
    }
    // Constructor to convert a
    // string into binary format.
    BitInteger(string str)
    {
        int len = str.length();
        int x = INTEGER_SIZE - len;
        bits = new bool[INTEGER_SIZE];
 
        // If len = 4. Then x = 32 - 4 = 28.
        // Hence iterate from
        // bit 28 to bit 32 and just
        // replicate the input string.
        int i = 0;
 
        for (int j = x; j <= INTEGER_SIZE && i < len; j++, i++) {
            if (str[i] == '1')
                bits[j] = true;
            else
                bits[j] = false;
        }
    }
 
    // this function fetches the kth bit
    int fetch(int k)
    {
        if (bits[k])
            return 1;
 
        return 0;
    }
 
    // this function will set a value
    // of bit indicated by k to given bitValue
    void set(int k, int bitValue)
    {
        if (bitValue == 0)
            bits[k] = false;
        else
            bits[k] = true;
    }
 
    // convert binary representation to integer
    int toInt()
    {
        int n = 0;
        for (int i = 0; i < INTEGER_SIZE; i++) {
            n = n << 1;
            if (bits[i])
                n = n | 1;
        }
        return n;
    }
};
 
// Function to find the missing number
int findMissingFunc(list<BitInteger>& myList, int column)
{
    // This means that we have processed
    // the entire 32 bit binary number.
    if (column < 0)
        return 0;
 
    list<BitInteger> oddIndices;
    list<BitInteger> evenIndices;
 
    for (BitInteger t : myList) {
 
        // Initially column = LSB. So
        // if LSB of the given number is 0,
        // then the number is even and
        // hence we add it to evenIndices list.
        // else if LSB = 0 then add it to oddIndices list.
        if (t.fetch(column) == 0)
            evenIndices.push_back(t);
        else
            oddIndices.push_back(t);
    }
 
    // Step 1 and Step 2 of the algorithm.
    // Here we determine the LSB bit of missing number.
 
    if (oddIndices.size() >= evenIndices.size())
 
        // LSB of the missing number is 0.
        // Hence it is an even number.
        // Step 3 and 4 of the algorithm
        // (discarding all odd numbers)
        return (findMissingFunc(evenIndices, column - 1)) << 1 | 0;
 
    else
        // LSB of the missing number is 1.
        // Hence it is an odd number.
        // Step 3 and 4 of the algorithm
        // (discarding all even numbers)
        return (findMissingFunc(oddIndices, column - 1)) << 1 | 1;
}
 
// Function to return the missing integer
int findMissing(list<BitInteger>& myList)
{
    // Initial call is with given array and LSB.
    return findMissingFunc(myList, BitInteger::INTEGER_SIZE - 1);
}
 
// Driver Code.
int main()
{
 
    // a corresponds to the input array which
    // is a list of binary numbers
    list<BitInteger> a = { BitInteger("0000"), BitInteger("0001"),
                           BitInteger("0010"), BitInteger("0100"),
                           BitInteger("0101") };
    int missing1 = findMissing(a);
    cout << missing1 << "\n";
 
    return 0;
}

#include <bitset>
#include <iostream>
#include <string>
#include <vector>

int find_missing_number(const std::vector<std::string>& arr)
{
    int n = arr.size(); // The number of binary strings in
                        // the array. Since one number is
                        // missing, n = N
    int xor_all = 0;

    // XOR all integers from 0 to N (inclusive)
    for (int i = 0; i <= n; ++i) {
        xor_all ^= i;
    }

    int xor_arr = 0;

    // XOR all the integers represented by the binary
    // strings in the array
    for (const auto& num : arr) {
        xor_arr ^= std::bitset<32>(num).to_ulong();
    }

    // The missing number is the XOR of xor_all and xor_arr
    int missing_number = xor_all ^ xor_arr;
    return missing_number;
}

int main()
{
    std::vector<std::string> arr
        = { "0000", "0001", "0010", "0100" };
    std::cout << find_missing_number(arr) << std::endl;
    return 0;
}

// This code is contributed by Shivam