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DAA Quiz 1

Authored by Aksheya Suresh

Computers

University

Used 26+ times

DAA Quiz 1
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10 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is time complexity of fun()?

int fun(int n)

{

int count = 0;

for (int i = n; i > 0; i /= 2)

for (int j = 0; j < i; j++)

count += 1;

return count;

}

O(n2)

O(nLogn)

O(n)

O(nLognLogn)

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the time complexity of fun()?

int fun(int n)

{

int count = 0;

for (int i = 0; i < n; i++)

for (int j = i; j > 0; j--)

count = count + 1;

return count;

}

Theta (n)

Theta (n2)

Theta (nLogn)

Theta (nLognLogn)

3.

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

The recurrence relation capturing the optimal time of the Tower of Hanoi problem with n disc’s is

T(n)=2T(n–2)+2

T(n) = 2T(n – 1) + n

T(n)=2T(n/2)+1

T(n) = 2T(n – 1) + 1

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Let w(n) and A(n) denote respectively, the worst case and average case running time of an algorithm executed on an input of size n. which of the following is ALWAYS TRUE?

A(n)=Omega(W(n))

A(n)=Theta(W(n))

A(n)=O(W(n))

None of these

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which of the given options provides the increasing order of asymptotic complexity of functions f1, f2, f3 and f4?

f1(n) = 2^n

f2(n) = n^(3/2)

f3(n) = nLogn

f4(n) = n^(Logn)

f3, f2, f4, f1

f3, f2, f1, f4

f2, f3, f1, f4

f2, f3, f4, f1

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

What is the best case time complexity of fun()?

void fun(int n, int arr[])

{

int i = 0, j = 0;

for(; i < n; ++i)

while(j < n && arr[i] < arr[j])

j++;

}

O(n)

O(n2)

O(nlogn)

O(n(logn)2)

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

In a competition, four different functions are observed. All the functions use a single for loop and within the for loop, same set of statements are executed. Consider the following for loops, If n is the size of input(positive), which function is most efficient (if the task to be performed is not an issue)?

for(i = 0; i < n; i++)

for(i = 0; i < n; i += 2)

for(i = 1; i < n; i *= 2)

for(i = n; i > -1; i /= 2)

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