Question

For the following program fragment give a Big-O analysis of the running time. Briefly explain your answer:

int t = 0;

for(int i=1; i <= n; i++)

for(int j=1; j <= i*i; j++)  

if(j % i == 0)   

t++;   

What I have so far, O(1) + O(n) + O(n²) + O(1) + O(1)

Drop Low order terms: O(n) + O(n²)

And I believe the final answer to be O(n³), but not sure if just drop the O(n) or keep it.

Thanks

Answer 1

The time complexity of each line is O(1) + O(n) + + O(1) + O(1).

The third line(due to being nested in second line) runs total of which is approx. equal to .

Big-O complexity of a code segment is always equal to maximum of the time complexity of each line.

So,time complexity of this program fragment is max(O(1) , O(n) , , O(1) , O(1)) which is equal to .

This is the logic behind finding the Big- O time complexity of a program segment.

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