GIVEN AN ARRAY OF INTEGERS OF SIZE N WHICH CONTAINS NUMBERS FROM 0 TO N – 2. EACH NUMBER IS PRESENT AT LEAST ONCE. THAT IS, IF N = 5, NUMBERS FROM 0 TO 3 IS PRESENT IN THE GIVEN ARRAY AT LEAST ONCE AND ONE NUMBER IS PRESENT TWICE. YOU NEED TO FIND AND RETURN THAT DUPLICATE NUMBER PRESENT IN THE ARRAY.

#include <iostream>
using namespace std;
// arr - input array
// size - size of array

int DuplicateNumber(int arr[], int size){
    /* Don't write main().
     * Don't read input, it is passed as function argument.
     * Return output and don't print it.
     * Taking input and printing output is handled automatically.
     */
    int sum=0;
    for(int i=0;i<size;i++)
    {
        sum+=arr[i];
    }
    int sumnatural=((size-1)*(size-2))/2;
    int duplicate=sum-sumnatural;
    return duplicate;
}

int main() {
	int size;
	cin >> size;
	int *input = new int[1 + size];
	
	for(int i = 0; i < size; i++)
		cin >> input[i];
	
	cout << DuplicateNumber(input, size);	
	
	delete [] input;

	return 0;
}