Expected time complexity of O(n^2), but it results in O(n). Can some explain why?

Question

What should be the time complexity of the following code? I tried to think and come up with O(n ² ) but the output says it to be of O(n). Can someone please explain through code?

for(int i = 0; i < n; i++){
    for(; i < n; i++){
        cout << i << endl;
    }
}

Answer 1

The complexity of your code is O(n).

Why?

Because, even though you have written two for loops, which probably made you thinking the complexity is O(n ² ), your code is actually one for loop like:

for (i = 0; i < n; i++){
    std::cout << i << std::endl;
}

Once the inner for loop finishes, i is equal to n and therefore the condition of outer for loop i < n is no longer satisfied.

Answer 2

One point to be noted while using such for loops is that your using a single variable.

Irrespective of how many outer loops you add, your code will result in the same with the condition i<n prevailing in all. The innermost loop is the one which will run till i=n-1 , the rest of which simply won't satisfy the condition.

for(int i=0; i<n; i++)
{ for(; i<n; i++)
  { for(; i<n; i++)
    { for(; i<n; i++) // and so on.
       std::cout<<i<<"\n";
    }
  }
}

Providing a variant to this, if you were to observe one such case of O(n ² ) complexity, your condition would have been i<n*n :

for(int i=0; i<n; i++)
{ for(; i<n*n; i++)
    std::cout<<i<<"\n";
}

Answer 3

Time complexity of your code is O(n) and not O(n^2) because when inner loop ends, at that time value of i has already reached to n. So outer loop cannot run any more.

for(int i = 0; i < 2; i++){
    for(; i < 2; i++){
        cout << i << endl;
    }
   //after loop run two times i has value 2.
   //and outer loop cannot run anymore
}