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GATE | GATE CS 1996 | Question 38

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  • Difficulty Level : Easy
  • Last Updated : 03 Jul, 2020
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The average number of key comparisons done in a successful sequential search in a list of length it is
(A) log n
(B) (n-1)/2
(C) n/2
(D) (n+1)/2

Answer: (D)

Explanation: If element is at 1 position then it requires 1 comparison.
If element is at 2 position then it requires 2 comparison.
If element is at 3 position then it requires 3 comparison.
Similarly , If element is at n position then it requires n comparison.

Total comparison 
= n(n+1)/2

For average comparison 
= (n(n+1)/2) / n 
= (n+1)/2 

Option (D) is correct.

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