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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 / 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 |
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