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ipn [44]
3 years ago
7

An array of integers named parkingTickets has been declared and initialized to the number of parking tickets given out by the ci

ty police each day since the beginning of the current year. (Thus, the first element of the array contains the number of tickets given on January 1; the last element contains the number of tickets given today.) A variable named mostTickets has been declared, along with a variable k. Without using any additional variables, write some code that results in mostTickets containing the largest value found in parkingTickets.
Computers and Technology
1 answer:
Sedaia [141]3 years ago
6 0

Answer:

Following code will store the largest value in array parkingTickets in the variable mostTickets

mostTickets = parkingTickets[0];

for(int k = 0; k<parkingTickets.length; k++)

{

if(parkingTickets[i]>mostTickets)

{

 mostTickets = parkingTickets[i];

}

}

Explanation:

In the above code segment, initially the number of tickets at first index is assumed as largest value of tickets in array.

Then using a for loop each value in the array parkingTickets is compared with the current mostTickets value.

If the compared value in parkingTickets array is larger than the current mostTickets value. Then that value is assigned to mostTickets.

This process is repeated for all elements in array.

Thus after looping through each element of array the largest value in array will get stored in mostTickets variable.  

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Part b: If the list contains n elements (where n is even) the middle terms are  at index (n-2)/2 & n/2 and the number of comparisons are (n+2)/2.

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there are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(9,10,11,12,13,14,15,16) above it.It will have to run 9 comparisons  to find the middle term.

If list has 21 elements, the middle item would be given at 10th index i.e.

there are 10 indices (0,1,2,3,4,5,6,7,8,9) below it and 10 indices (11,12,13,14,15,16,17,18,19,20) above it.It will have to run 11 comparisons  to find the middle term.

Now this indicates that if the list contains n elements (where n is odd) the middle term is at index (n-1)/2 and the number of comparisons are (n+1)/2.

Part b

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If list has 18 elements, there are two middle terms as  one at would be at 8th index and the one at 9th index .There are 8 indices(0,1,2,3,4,5,6,7) below it and 8 indices(10,11,12,13,14,15,16,17) above it. It will have to run 10 comparisons  to find the middle terms.

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Part c

So the average number of comparisons is given as

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So the average number of comparisons for a list bearing n elements is 2n+3/4 comparisons.

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