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Otrada [13]
2 years ago
8

Consider two vectors that are NOT sorted, each containing n comparable items. How long would it take to display all items (in an

y order) which appear in either the first or second vector, but not in both, if you are only allowed LaTeX: \Theta\left(1\right)Θ ( 1 ) additional memory? Give the worst-case time complexity of the most efficient algorithm.
Computers and Technology
1 answer:
Darina [25.2K]2 years ago
7 0

Answer:

The correct answer to the following question will be "Θ(​n​2​)

". The further explanation is given below.

Explanation:

If we're to show all the objects that exist from either the first as well as the second vector, though not all of them, so we'll have to cycle around the first vector, so we'll have to match all the objects with the second one.

So,

This one takes:

= O(n^2)

And then the same manner compared again first with the second one, this takes.

= O(n^2)

Therefore the total complexity,

= Θ(​n​2​)

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Which of the following is an object such as a field which can be inserted into a document
ivanzaharov [21]

Answer:

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Explanation:

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8 0
2 years ago
Using the college registration example from section 6.7.3 as a starting point, do the following:
Arada [10]

Answer:

Explanation:

NCLUDE Irvine32.inc

TRUE = 1

FALSE = 0

.data

gradeAverage WORD ?

credits WORD ?

oKToRegister BYTE ?

str1 BYTE "Error: Credits must be between 1 and 30" , 0dh,0ah,0

main PROC

call CheckRegs

exit

main ENDP

CheckRegs PROC

push edx

mov OkToRegister,FALSE

; Check credits for valid range 1-30

cmp credits,1 ; credits < 1?

jb E1

cmp credits,30 ; credits > 30?

ja E1

jmp L1 ; credits are ok

; Display error message: credits out of range

E1:

mov edx,OFFSET str1

call WriteString

jmp L4

L1:

cmp gradeAverage,350 ; if gradeAverage > 350

jna L2

mov OkToRegister,TRUE ; OkToRegister = TRUE

jmp L4

L2:

cmp gradeAverage,250 ; elseif gradeAverage > 250

jna L3

cmp credits,16 ; && credits <= 16

jnbe L3

mov OkToRegister,TRUE ; OKToRegister = TRUE

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L3:

cmp credits,12 ; elseif credits <= 12

ja L4

mov OkToRegister,TRUE ; OKToRegister = TRUE

L4:

pop edx ; endif

ret

CheckRegs ENDP

END main

8 0
2 years ago
You need to replace a broken monitor on a desktop system. You decide to replace it with a spare monitor that wasn't being used.
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Firmware hope this helps!!
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Use the loop invariant (I) to show that the code below correctly computes the product of all elements in an array A of n integer
NeTakaya

Answer:

Given Loop Variant P = a[0], a[1] ... a[i]

It is product of n terms in array

Explanation:

The Basic Step: i = 0, loop invariant p=a[0], it is true because of 'p' initialized as a[0].

Induction Step: Assume that for i = n - 3, loop invariant p is product of a[0], a[1], a[2] .... a[n - 3].

So, after that multiply a[n - 2] with p, i.e P = a[0], a[1], a[2] .... a[n - 3].a[n - 2].

After execution of while loop, loop variant p becomes: P = a[0], a[1], a[2] .... a[n -3].a[n -2].

for the case i = n-2, invariant p is product of a[0], a[1], a[2] .... a[n-3].a[n-2]. It is the product of (n-1) terms. In while loop, incrementing the value of i, so i=n-1

And multiply a[n-1] with p, i.e P = a[0].a[1].a[2].... a[n-2].

a[n-1]. i.e. P=P.a[n-1]

By the assumption for i=n-3 loop invariant is true, therefore for i=n-2 also it is true.

By induction method proved that for all n > = 1 Code will return product of n array elements.

While loop check the condition i < n - 1. therefore the conditional statement is n - i > 1

If i = n , n - i = 0 , it will violate condition of while loop, so, the while loop will terminate at i = n at this time loop invariant P = a[0].a[1].a[2]....a[n-2].a[n-1]

6 0
2 years ago
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