Question

int count = 0; for (int i = 0; i < N; i++) { if (unlucky) { for (j = N; j > i; j--) { count = count + i + j; } } }

          int count = 0;
for (int i = 0; i < N; i++)
{
if (unlucky)
{
for (j = N; j > i; j--)
{
count = count + i + j;
}
}
}

Added by Dennis Y.

Computer Science and Information Technology

Trishna Knowledge Systems 2018 Edition

Instant Answer

Solved by Expert Tarandeep Singh

Step 1

1) Using operation counting, we can count the number of operations for different cases: - Best case: In the best case, the condition "if (unlucky or = Nj > ij--)" is never true, so the loop is not executed. Show more…

Show all steps

Thanks for your feedback!

For the above algorithm: 1) Using operation counting, what are the number of operations for: - The best case - The worst case - The average case 2) Using Big O notation, describe: - The best case - The worst case - The average case 3) Describe the algorithm's overall performance using Big-O. 4) Describe the algorithm's overall performance using Big-â„¦. 5) Describe the algorithm's overall performance using Big-Theta. 6) Select which of the above asymptotic notations is the best way to describe the algorithm's performance. int: count: :: 0: foriint=OiNi++ ifunlucky or=Nj>ij--) 1 count=coun+i+j