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 / saurav
i dont think the above answer is correct..it should be (N+1)/2
according to me..if my answer is not correct,can anyone
explain me the reason..
| Is This Answer Correct ? | 54 Yes | 4 No |
Answer / wonder
Avarage number of comparision will be (N+1)/2(N-size of
array).
Because:If elements is in 1st position no of cpmparision
will be one and if the element is in the last position then
no of comparisions will be N.
| Is This Answer Correct ? | 31 Yes | 0 No |
Answer / nilavalagan
The element may be found at any place in the array.
Supposing that the element is in 1st position, no.of comparison = 1;
Element is in 2nd position, no.of comparison = 2;
In the same way,
for the element present in 100th positionn no.of comparison = 100;
Total no.of comparison = 1+2+3+....+100
Average no.of comparison = (1+2+3+..+100)/100
= (100*101)/2*100
(Remember n(n+1)/2)
=101/2 = 50.5
In general = (n+1)/2; if there are n elements.
| Is This Answer Correct ? | 15 Yes | 0 No |
Answer / shyam
see
http://www.cs.odu.edu/~cmo/classes/msim602/studyGuide/ch10.doc.
for details..
the fact that the array is already sorted does not affect
the number of comparisons ..
| Is This Answer Correct ? | 6 Yes | 2 No |
Answer / dj
Answer is log n (base 2), Maximum camparision required to
search the array of 100 elments is 8. ( modified binary
search)
| Is This Answer Correct ? | 10 Yes | 7 No |
Answer / sherin
its asked sequential search
if it was binary then dividing array into 2 , then
comparing....100 - 50 - 25 - 12 - 6 - 3 - 1 -then the no.,
i think Dj thought it like that ..
Sequential search is not needed here because the numbers are
ordered , if it was not ordered then 'n' comparisons are
necessary.
question here is not worst case scenario , see its asked
average case ..that too with ordered numbers ...
| Is This Answer Correct ? | 4 Yes | 1 No |
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