C++ Program to Find lost element from a duplicated array

// C++ program to find missing element from same
// arrays (except one missing element)
#include <bits/stdc++.h>
using namespace std;

// Function to find missing element based on binary
// search approach.  arr1[] is of larger size and
// N is size of it.  arr1[] and arr2[] are assumed
// to be in same order.
int findMissingUtil(int arr1[], int arr2[], int N)
{
    // special case, for only element which is
    // missing in second array
    if (N == 1)
        return arr1[0];

    // special case, for first element missing
    if (arr1[0] != arr2[0])
        return arr1[0];

    // Initialize current corner points
    int lo = 0,  hi = N - 1;

    // loop until lo < hi
    while (lo < hi)
    {
        int mid = (lo + hi) / 2;

        // If element at mid indices are equal
        // then go to right subarray
        if (arr1[mid] == arr2[mid])
            lo = mid;
        else
            hi = mid;

        // if lo, hi becomes contiguous,  break
        if (lo == hi - 1)
            break;
    }

    // missing element will be at hi index of
    // bigger array
    return arr1[hi];
}

// This function mainly does basic error checking
// and calls findMissingUtil
void findMissing(int arr1[], int arr2[], int M, int N)
{
    if (N == M-1)
        cout << "Missing Element is "
        << findMissingUtil(arr1, arr2, M) << endl;
    else if (M == N-1)
        cout << "Missing Element is "
        << findMissingUtil(arr2, arr1, N) << endl;
    else
        cout << "Invalid Input";
}

// Driver Code
int main()
{
    int arr1[] = {1, 4, 5, 7, 9};
    int arr2[] = {4, 5, 7, 9};

    int M = sizeof(arr1) / sizeof(int);
    int N = sizeof(arr2) / sizeof(int);

    findMissing(arr1, arr2, M, N);

    return 0;
}

// C++ program to find missing element from one array
// such that it has all elements of other array except
// one.  Elements in two arrays can be in any order.
#include <bits/stdc++.h>
using namespace std;

// This function mainly does XOR of all elements
// of arr1[] and arr2[]
void findMissing(int arr1[], int arr2[], int M,
                 int N)
{
    if (M != N-1 && N != M-1)
    {
        cout << "Invalid Input";
        return;
    }

    // Do XOR of all element
    int res = 0;
    for (int i=0; i<M; i++)
       res = res^arr1[i];
    for (int i=0; i<N; i++)
       res = res^arr2[i];

    cout << "Missing element is " << res;
}

// Driver Code
int main()
{
    int arr1[] = {4, 1, 5, 9, 7};
    int arr2[] = {7, 5, 9, 4};

    int M = sizeof(arr1) / sizeof(int);
    int N = sizeof(arr2) / sizeof(int);

    findMissing(arr1, arr2, M, N);

    return 0;
}