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Min-Max
Write an algorithm that finds both the smallest and
largest numbers in a list of n numbers and with complexity
T(n) is at most about (1.5)n comparisons.
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Answer Posted / amirhosain shahsavari
hi
Some of notations in answer 5 is absolutely wrong:
for even number n: T(n)=(3/2)n-2
for odd number n: we ignore the last item of array.Now you
can find Min1 (minimum of the array items ignoring the last
item), Max1 (maximum of the array items ignoring the last
item) (notice that n-1 is even so Min1 and Max1 can be found
at (3/2)(n-1)-2 comparisons). After that you should compare
Min1 and Max1 with the last item (that you ignored in the
first step). so you need at most (3/2)(n-1)-2+2=(3/2)n-(3/2)
comparisons to find the entire array items
| Is This Answer Correct ? | 2 Yes | 1 No |
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