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POBest case, average case, worst case, and amortized case analysis
 

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  1. POBest case, average case, worst case, and amortized case analysis
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    <p>I am looking of an understanding of how things work, not just the answer. I'm giving what I've though the answer is, but I'm not really sure and the text doesn't have much on this kind of analysis.</p> <p>I am quite aware that this NOT how you are supposed to implement a stack, but this is the scenario we have been given.</p> <p>Part A</p> <blockquote> <p>What is the worst case, average case, and best case running time of insertion into slot a[0] of an (unsorted) array a containing N integers, representing a stack? Give theta bounds. </p> </blockquote> <p>Part B</p> <blockquote> <p>What is the amortized running time of insertion of N integers into an initially empty array a, representing a stack, if every insertion takes place at a[0]? Give a tight O bound and explain your answer.</p> </blockquote> <p>The instructor was quite clear that we are doing all insertions into a[0], and there are no removals from this "stack".</p> <p>My analysis is this</p> <p>Assumption: a.length >> N</p> <p>Best case is Theta(1) which occurs when we insert into an empty "stack".</p> <p>Worst Case is Theta(N) because as part of the insertion process we have to shift N-1 elements to make room at a[0].</p> <p>Average case is also Theta(N) because, no matter what, we always have to shift N-1 elements.</p> <p>Amortized case</p> <p>Each insertion costs (N-1) and we have N elements, so we use:</p> <pre><code>N Sum (i) = [ N(N+1) ] / 2 = N^2 / 2 which is O(N^2) for N insertions i=1 </code></pre> <p>Amortized cost would then be N^2 / N = O(N) </p> <p>Problem I've having is that this looks to easy. Am I missing something, or do I have it pretty much dead on?</p>
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    1. USGeldhart
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    1. COI think you're probably right, except that the "best case" in part A is still Theta(N). "Best case" means the best, *for the size N given*, it doesn't mean for the best (i.e. smallest) value of N. So what's weird here, is that you're being asked best/worst/average for an operation that's not data-dependent. It's possible that you aren't supposed to assume `a.length >> N`, rather the array is re-allocated whenever its size is insufficient. But I think that's unlikely, there's nothing in the text you've quoted to indicate that we're supposed to understand an "array" to mean that.
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      1. POBest case, average case, worst case, and amortized case analysis
      1. USSteve Jessop
    2. COThere is something strange in your assumption, that seems to have been overlooked, and maybe there's a trick, an escape window that the instructor really wanted you to find. I won't tell you yet, but I'll give a hint as a question: on insertion, how would you keep track of the current number of inserted elements (the i that you assume available in your calculation)?
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      1. USslartibartfast
 

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