(Solved):What is the Big-Oh order of the following code fragment? The size of the problem is expressed as n. for (int i = 0; i < (int)Math.pow(2, n); i++) System.out.println("what could could go wrong?");... View Answer...

 

Question

What is the Big-Oh order of the following code fragment? The size of the problem is expressed as n.

for (int i = 0; i < (int)Math.pow(2, n); i++)
System.out.println("what could could go wrong?"); //f(n) counts these

 

EXPERT ANSWER

Here the loop will run {eq}2^n

{/eq} times depending on the value of n. So the the Big-Oh Order will be {eq}O(2^n)

{/eq}

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