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. ~ advance coding in data structures and algorithms c++

TITLE:-

Duplicate in array

PROBLEM:-

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.

Assume, duplicate number is always present in the array.

Input format :

Line 1 : Size of input array
Line 2 : Array elements (separated by space)

Output Format :

Duplicate element

Constraints :

1 <= n <= 10^6

Sample Input:

9
0 7 2 5 4 7 1 3 6

Sample Output:

SOLUTION:-

#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;
}

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