What is the average number of comparisons in a sequential
search?
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Answer / vivek tiwari
f(n)= 1.Pn + 2.Pn + 3.Pn +...+ N.Pn
where
Pn = 1/N
f(n)= 1.1/N +2.1/N + 3.1/N+....+N.1/N
= (1+2+3+....+N)1/N
= N(N+1)/2N
= (N+1)/2
| Is This Answer Correct ? | 47 Yes | 1 No |
Answer / bipin from utkal university mc
suppose there are five element 23,56,78,12,90
minimum time require means searching element present at
first so it takes only one comparison
maximum time require means searching element present at
last so it takes n No. of comparison(here 5 )
so avarage comarison=(1+n)/2
| Is This Answer Correct ? | 10 Yes | 1 No |
Answer / kalyani
Searching an element in an array, the search starts from the first element till the last element the average number of comparisons in a sequential search is (N+1)/2,where N is the size. The number of comparisons will be 1, if element is in the first position and if element is in the last position the number of comparisons will be N.
| Is This Answer Correct ? | 0 Yes | 0 No |
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