Is this code correct in place of the one in video?

#1

int main ()
{
int n, i;
cin >> n;

	for(i=2;i<n;i++)	
	{
		if(n%i==0)	
			{
				cout<<"Not Prime" << endl;
				break;
			}
			
	}
	
if(n==i)	
	{
		cout<<"Prime"<<endl;
		
	}
	
	
	
	return 0;
	
}
#2

Yes this is correct to find whether a number is prime or not but the time complexity is O(N)

#3

Sorry, I didnt get the time complexity part. What do you mean when you say “the time complexity is O(N)”.

#4

Hey Divyam, time complexity is O(N) as you are traversing from 2 to N once. If you haven’t studied the time complexity till now, do watch the videos given in the course in time and space complexity section.

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