FindDuplicate.cpp

// You have been given an integer array/list(ARR) of size N which contains numbers from 0 to (N - 2). 
// Each number is present at least once. That is, if N = 5, the array/list constitutes values ranging from 0 to 3 and among these, there is a single integer value that is present twice. 
// You need to find and return that duplicate number present in the array.
// Note : Duplicate number is always present in the given array/list.

// My Code
int duplicateNumber(int *arr, int size)
{
    int count, res;
    for(int i=0; i<size; i++){
        int j= arr[i];
        count = 0;
        for(int k=0; k<size; k++){
            if (arr[k] == j){
                count++;
            }
        }
        if(count > 1){
            res = arr[i];
        }
    }
    return res;
}

// Main Code
#include <iostream>
using namespace std;

#include "solution.h"

int main()
{

	int t;
	cin >> t;
	
	while (t--)
	{
		int size;
		cin >> size;
		int *input = new int[size];

		for (int i = 0; i < size; i++)
		{
			cin >> input[i];
		}

		cout << duplicateNumber(input, size) << endl;
	}

	return 0;
}

// CN Code
int duplicateNumber(int *arr, int size)
{
    for(int i=0; i<size; i++){
        for(int j=i+1; j<size; j++){
            if(arr[i] == arr[j]){
                return arr[i];
            }
        }
    }
}