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yan [13]
4 years ago
9

(20 points) Given a collection of n nuts and a collection of n bolts, arranged in an increasing order of size, give an O(n) time

algorithm to check if there is a nut and a bolt that have the same size. The sizes of the nuts and bolts are stored in the sorted arrays NUT S[1..n] and BOLT S[1..n], respectively. Your algorithm can stop as soon as it finds a single match (i.e, you do not need to report all matches).
Computers and Technology
1 answer:
Blababa [14]4 years ago
6 0

Answer:

Check the explanation

Explanation:

Keep two iterators, i (for nuts array) and j (for bolts array).

while(i < n and j < n) {

if nuts[i] == bolts[j] {

We have a case where sizes match, output/return

}

else if nuts[i] < bolts[j] {

what this means is that the size of nut is lesser than that of bolt and we should go to the next bigger nut, i.e., i+=1

}

else {

what this means is that the size of bolt is lesser than that of nut and we should go to the next bigger bolt, i.e., j+=1

}

}

Since we go to each index in both the array only once, the algorithm take O(n) time.

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Check the explanation

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(1.) The functional dependencies based on assumptions are as follows:-

Trip determines which staff was there in the trip, the car used for the trip, the quantity of fuel used in the trip, the fill up time and the total paid of the trip.

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(2.) Initially the relation is as follows:-

R (Trip ID, Staff Name, Car Details, License Plate, Odometer Reading, Service Station, Station Address, Fill up Time, Fuel type, Quantity, Cost per Litre, Total Paid)

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R2 (Service Station, Street, City, State, Zip Code)

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