Checking duplicated numbers in Array - c++

Question

The function should check if there is any duplicated number but it's not working somehow.
anyone knows why?
#include <iostream>
using namespace std;
int main()
{
int const size = 10;
int arry[size] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 ,10};
int i = 0;
int k = 1;
for (i = 0; i < size; i++)
{
for (k = 1; k < size; k++)
{
if (arry[i] == arry[k])
{
cout << arry[i] << " " << arry[k] << endl;
cout << "duplicate" << endl;
return 0;
}
}
}
cout << "no duplicate" << endl;
return 0;
}

Answer 1

You should skip when i == k.
for (i = 0; i < size; i++)
{
for (k = i+1; k < size; k++) // <-- either change HERE
{
if (i != k && arry[i] == arry[k])
// ^^^^^^ or HERE
{
cout << arry[i] << " " << arry[k] << endl;
cout << "duplicate" << endl;
return 0;
}
}
}

Answer 2

You're checking every combination twice - for example, you're comparing element 2 with element 5, and then 5 with 2. Aside from the fact that you're also checking an element with itself (which is what timrau's answer is guarding against), you're doing the same work twice.
So try
for (i = 0; i < size; i++)
{
for (k = i + 1; k < size; k++)
instead.
Your approach still doesn't scale very well, it's essentially O(n^2). It would be faster to sort your array and then look for duplicate neighbors.

Answer 3

A more efficient approach should be to sort the array first:
std::vector<int> a = {4, 3, 2, 1, 2, 5, 6};
std::sort(std::begin(a), std::end(a));
for (std::size_t i = 1; i < a.size(); ++i) {
if (a[i - 1] == a[i]) {
std::cout << a[i] << " " << a[i] << std::endl;
std::cout << "duplicate" << std::endl;
return 0;
}
}
return 0;