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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 / 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 |
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