Question

Analyze the following code fragments and write down the Big-O estimates of the following code fragments. Provide a concise explanation how you got your answer.

c. for (int j = 0; j < n; j++)

{

    for (int k = 0; k < n; k++)

         cout << (j + k) << endl;

}

d. while (n > 1)

{   

k += n *3;

n = n / 2;

}

e. int temp = n;

for (int j = 0; j < n; j++)

{

while (temp > 1)  

temp = temp / 2;

}

Answer 1

c.

for (int j = 0; j < n; j++)

{

for (int k = 0; k < n; k++;

cout << (j + k) << endl;

}

Answer: O(n²)

Explanation:

1. The given method has nested for loop.

2. The inner loop executes n times and the outer loop also executes n times, Hence the total complexity isO(n²).

d.

while (n > 1)

{   

k += n *3;

n = n / 2;

}

Answer: O(n/2)

Explanation:

Complexity is considered to be the worst case implementation, Let us consider the n = 2; Then the while loop is executed 1 time;

Let the n value is 8; then the loop is executed 3 times; As the value of n increses the number times the loop execution decreases, but when compared to the worst case scenario, the maximum times the statements executed is n/2;

So the complexity is O(n/2).

e.

int temp = n;

for (int j = 0; j < n; j++)

{

while (temp > 1)  

            temp = temp / 2;

}

Answer: O(n)

Explanation:

The inner executes n/2 as explained in the part e. similarly the outer loop executes n times. So the complexity is O(n, n/2). The maximum of two is n. Therefore the complexity is O(n).