Need Big O rating for the 4 groups of code if (n < 100) System.out.println (\"n:
ID: 3546669 • Letter: N
Question
Need Big O rating for the 4 groups of code
if (n < 100)
System.out.println ("n: " + n); O(1)
for (int x = 0; x < n % 2; x++)
System.out.println ("x: " + x);
for (int x = 0; x < n; x++)
for (int j = 0; j < m; j++)
System.out.println ("x: " + x); O(n^2)
for (int x = 0; x < n; x++)
for (int j = 0; j < n; j++)
if (x < 0)
for (int k = 0; k < n % 10; k++)
System.out.println ("x: " + x);
if (n < 100)
System.out.println ("n: " + n); O(1)
for (int x = 0; x < n % 2; x++)
System.out.println ("x: " + x);
for (int x = 0; x < n; x++)
for (int j = 0; j < m; j++)
System.out.println ("x: " + x); O(n^2)
for (int x = 0; x < n; x++)
for (int j = 0; j < n; j++)
if (x < 0)
for (int k = 0; k < n % 10; k++)
System.out.println ("x: " + x);
Explanation / Answer
if (n < 100)
System.out.println ("n: " + n); O(1) as it is in one single statement and execute once so it take constant time
for (int x = 0; x < n % 2; x++)
System.out.println ("x: " + x); O(n) as it run until x < n%2 which can run some n/constant time but on removing
the constant coofficent it running time in O(n)
for (int x = 0; x < n; x++)
for (int j = 0; j < m; j++)
System.out.println ("x: " + x); O(n^2) as first for loop run second for loop n time and the second for loop run n time
and the System.out.println ("x: " + x); run n^2 hence it run O(n^2)
for (int x = 0; x < n; x++)
for (int j = 0; j < n; j++)
if (x < 0)
for (int k = 0; k < n % 10; k++)
System.out.println ("x: " + x); O(n^3) as first for loop run second for loop n time and the second for loop run n time
third for loop n time System.out.println ("x: " + x); run n^3 hence it run
O(n^3)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.