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What is the average number of comparisons needed in a
sequential search to determine the position of an element in
an array of 100 elements, if the elements are ordered from
largest to smallest?
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Answer Posted / gianni
the above answer of n+1/2 is correct if you assume it will
always be a successful search. you also need to take into
account the probability of the item NOT being in the list.
as such your final formula is actually
p/n * n(n+1)/2 + n(1-p) where p is the probability that the
item is in the list. That formula can be reduced ton(1-p/2)+p/2
assuming a probability of .5, you wind up with an answer of
(3n+1)/4 or ~3/4 of the list will be searched on average.
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