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POComputational complexity of a algorithm
 

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  1. POComputational complexity of a algorithm
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    <pre><code>Algorithm 1. STACKSTUFF(n) Input: Integer n 1) Let S = an empty Stack 2) Let X = -1 3) For i = 1 to 2n 4) S.Push(i) 5) End For 6) For i = 1 to n 7) Let X = S.Pop() 8) End For Output: The contents of X </code></pre> <hr> <pre><code>1) What is this algorithm written in pseudo code doing? </code></pre> <p>To my understanding, S.Push(i) adds item i on top of stack S. X = S.Pop() removes the item from the top of stack S and assigns it to X.</p> <pre><code>2) What is the computational complexity O(n) for algorithm 1, STACKSTUFF? </code></pre> <p>I believe the answer would be: O(3n)</p> <p>The first loop would be 2n and the second for loop n, so 2n+n=3n.</p> <p>Or... Would the answer just be O(n^2) since all we would have to do would be n*n?</p> <pre><code>3) If n &gt; 0 then what is returned by the algorithm? What about n &lt; 1 a) 2n b) -1 c) n-1 d) n+1 e) None of the above </code></pre> <p>This last bit really confuses me. From my understanding, if n was always greater than 0, the algorithm would always return n+1, and if n was always less than 1, the algorithm would return n-1. However this is pure guess work...</p> <p>If I thought about this logically, then let's say n was 3 for example. Since the first For loop is 1 to 2n, then this would mean that we would end up with the following stack S={1,2,3,4,5,6} as it added every number up to double n into S. The second For loop then pops 3 numbers so X ends up looking like this X={6,5,4}. If I am correct there... Should I assume that this was just a trick question and the answer is e, none of the above?</p> <p>I just wanted to make sure my understanding here was correct before I continued studying. Thanks for any help.</p>
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