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lawyer [7]
3 years ago
10

Using Sequential Search on an array of size n, the search key is definitely present in the array. The probability of matching th

e key to the nth item in the array is 1/3, and the probability of matching the key to the (n-1)st item is 1/2. The probabilities of matching each of the remaining items are all equal. What is the average case complexity function for Sequential Search under these conditions?
Computers and Technology
1 answer:
MissTica3 years ago
7 0

Answer:

(11n-5) / 12 is correct answer.

Explanation:

The Probability that key will match to nth term = 1/2

The Probability that key will match to n-1th term = 1/3

As all other probabilities are equal

The Total Probability that key matches to any of 1 to n-2 index = 1 - 1/2 - 1/3 = 1/6

The Probability that key matches to any of 1 to n-2 index = (1/6) / n-2 = (1/6)* (n-2))

Let P(i) = Probability that key matches to ith index.

The Average time complexity = 22 i=1 P(i) * i

The Average time complexity = 1/(6(n-2) * ( sum of 1 to n-2 ) + (n-1) / 3 + n/2

The Average time complexity = 1/(6(n-2) * ( n-2)*(n-1) / 2 + ( n-1) / 3 + n/2

The Average time complexity = 1/6 * (n-1)/2 + (n-1)/3 + n/2

The Average time complexity = (n-1)/12 + (n-1)/3 + n/2

The Average time complexity = (n-1 + 4 * n - 4 * 1 +6 * n)/12

The Average time complexity = 11n-5 / 12

so (11n-5) / 12 is correct answer.

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