Identify and state the running time using Big-O - algorithm

Question

For each of the following algorithms, identify and state the running time using Big-O.
//i for (int i = 0; Math.sqrt(i) < n; i++)
cout << i << endl;
//ii for (int i = 0; i < n; i++){
cout << i << endl;
int k = n;
while (k > 0)
{
k /= 2;
cout << k << endl;
} // while
}
//iii
int k = 1;
for (int i = 0; i < n; i++)
k = k * 2;
for (int j = 0; j < k; j++)
cout << j << endl;
I've calculate the loop times for the first question using n=1 and n=2. The loop in i will run n^2-1 times. Please help and guide me to identify the Big-O notation.

Answer 1

(i) for (int i = 0; Math.sqrt(i) < n; i++)
cout << i << endl;
The loop will run until squareRoot(i) < N , or until i < N^2. Thus the running time will be O(N^2), ie. quadratic.
(ii) for (int i = 0; i < n; i++){
cout << i << endl;
int k = n;
while (k > 0)
{
k /= 2;
cout << k << endl;
} // while
}
The outer loop will run for N iterations. The inner loop will run for logN iterations(because the inner loop will run for k=N, N/2, N/(2^2), N/(2^3), ...logN times). Thus the running time will be O(N logN), ie. linearithmic.
(iii)
int k = 1;
for (int i = 0; i < n; i++)
k = k * 2;
for (int j = 0; j < k; j++)
cout << j << endl;
The value of k after the execution of the first loop will be 2^n as k is multiplied by 2 n times. The second loop runs k times. Thus it will run for 2^n iterations. Running time is O(2^N), ie. exponential.

Answer 2

For the first question, you will have to loop until Math.sqrt(i) >= n, that means that you will stop when i >= n*n, thus the first program runs in O(n^2).
For the second question, the outer loop will execute n times, and the inner loop keeps repeatedly halving k (which is initially equal to n). So the inner loop executes log n times, thus the total time complexity is O(n log n).
For the third question, the first loop executes n times, and on each iteration you double the value of k which is initially 1. After the loop terminates, you will have k = 2^n, and the second loop executes k times, so the total complexity will be O(2^n)

Answer 3

Couple hints may allow you to solve most of running time complexity problems in CS tests/homeworks.
If something decrease by a factor of 2 on each iteration, that's a log(N). In your second case the inner loop index is halved each time.
Geometric series,
a r^0 + a r^1 + a r^2 ... = a (r^n - 1) / (r - 1).
Write out third problem:
2 + 4 + 8 + 16 ... = 2^1 + 2^2 + 2^3 + 2^4 + ...
and use the closed form formula.
Generally it helps to look for log2 and to write few terms to see if there is a repeatable pattern.
Other common questions require you to know factorials and its approximation (Sterling's approximation)

Answer 4

Using Sigma Notation, you can formally obtain the following results:
(i)
(ii)
(iii)