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What is the O-notation for the following code?

```
cin >> N;
for (int count = 1; count <= N; count++)
{
for (int count2 = 1; count2 <= count; count2++)
}
```

a) O (log N)

b) O (N)

c) O (N^2)

d) O (N^3)

outer loop is O(N)

and the inner loop is also O(N) so you multiply these together..

i'll get N*N = N^2 so the answer is "c" O(N^2)

but what if

```
cin >> N;
for (int count = 1; count <= N; count++)
{
for (int count2 = 1; count2 <= count; count2++)
{
cout << count1 + count2;
}
}
```

count1 = N

count2 = N

count1 + count2 = N + N??

so what is the answer in this case??...O(N) or O(N^2)???

a) O (log N)

b) O (N)

c) O (N^2)

d) O (N^3)