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Pseudocode Quiz 1

Question 1

Time: 00:00:00

Consider the below code and choose line can be replaced by which one of the following :

int sum (int A[], int n){

int sum = 0, i;

for (i = 0; i < n; i++) ------- line 1

sum = sum + A[i];

return sum;

}

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

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

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

for ( i = 0 ; i > n ; )

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++i and i++ both will increment i for next iteration in for loop.

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Question 2

Time: 00:00:00

What will be the output of the following pseudo code?

function (input a, input b)

  	if (a > b)

 		 return function (b, a)

  	elseif (b != 0)

  		return (a + function (a, b - 1))

  	else

  		return 0

function(6,8)

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When the function(6, 8) call since, a is not greater than b and b!=0 so else if condition will becomes true and it return 6 + function(6,7)...
Again when the function(6,7) will call it return (6+function(6,6)).. in this way a remains same and function will call by decrements the value of b when b =0 then it return 0 and recursively a will be added b times, hence we get output (6+6+6+6+6+6+6+6 = 48).

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Question 3

Time: 00:00:00

What will be the output of the following pseudo code?

Input m = 12, n = 4

m = m - 1 ;

n = n + 1 ;

m = m + n ;

n = n - 1 ;

Print m,n

16,4

15,4

4,16

17,4

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Input m = 12, n = 4
m = m - 1 ; // m = 11
n = n + 1 ; // n = 5
m = m + n ; // m = 16
n = n - 1 ; // 4
Print m,n // 16 4

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Question 4

Time: 00:00:00

What will be the output of the following pseudo-code?

Input i=8,j=9

Function ( i , j )

 for x=1 to x=5

	If i < j 

		i++

	Else 

		j++

print( i, j )

10,11

11,10

11,11

10,10

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Initially, i=8 and j=9
When x=1 , j>i // i = 9, j=9
When x=2, j is not greater than i //i=9, j=10
When x=3 , j>i // i = 10, j=10
When x=4, j is not greater than i //i=10, j=11
When x=3 , j>i // i = 11, j=11

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Question 5

Time: 00:00:00

What will be the time complexity of the following pseudo code?

Input i=8,j=9

Function ( i , j )

 for x=1 to x=5

	If i < j 

		i++

	Else 

		j++

print( i, j )

O(x)

O(1)

O(logx)

O(x^2)

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In the given code for loop will run for the value of x from 1 to 5, hence the complexity of the code will be O(x).

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Question 6

Time: 00:00:00

What will be the output of the following pseudo code?

function(int n)

   if (n <= 1)

      return n

   return function(n-1) + function(n-2)

End of function

main ()

   int n = 9

  printf("%d", function(n))

  return 0

End of main

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function(9) -> function(8)+function(7)
function(8) -> function(7)+function(6)
function(7) -> function(6)+function(5)
function(6) -> function(5)+function(4)
function(5) -> function(4)+function(3)
function(4) -> function(3)+function(2)
function(3) -> function(2)+function(1)
function(2) -> function(1)+function(0)
function(1) -> 1
function(0) -> 0
So, recursively, function(2) -> 1+0 => 1
function(3) => 1+1 =>2
function(4) => 2+1 => 3
function(5) => 2+3 => 5
function(6) => 5+3 => 8
function(7) => 8+5 => 13
function(8) => 13+8 => 21
function(9) => 21+13 => 34

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Question 7

Time: 00:00:00

What will be the time complexity of the following pseudo code?

function (int n)

{

    int a = 0, b = 1, c, i;

    if( n == 0)

    		return a;

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

    {

    		c = a + b;

    		a = b;

    		b = c;

    }

    return b;

}



int main ()

{

    int n = 9;

    printf("%d", function(n));

    return 0;

}

O(log(n))

O(n)

O(n^2)

O(1)

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The given pseudo code pseudo code represents the code to print the fibonacci series, so the complexity of the code will be O(n) as the for loop will run from 2 to n.

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Question 8

Time: 00:00:00

What does the following piece of code do?

public void func (Tree root)

{

      func (root.left ());

      func (root.right ());

      System.out.println (root.data ());

}

Pre-order

In-order

Level-order

Post-order

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The given piece of the code represents the Post-order tree Traversal
In this traversal method, the root node is visited last, hence the name. First we traverse the left subtree, then the right subtree and finally the root node.

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Question 9

Time: 00:00:00

How will you find the pre order traversal of a tree?

public void func (Tree root)
{
 func (root.left ());
 func (root.right ());
 System.out.println (root.data ());
}

public void func (Tree root)
{
 func (root.left ());
 func (root.right ());
 System.out.println (root.data ());
}

public void func (Tree root)
{
 System.out.println (root.data ());
 func (root.left ());
 func (root.right ());
}

public void func (Tree root)
{
 System.out.println (root.data ());
 func (root.left ());
 func (root.right ());
}

public void func (Tree root)
{
 func (root.left ());
 System.out.println (root.data ());
 func (root.right ());
}

public void func (Tree root)
{
 func (root.left ());
 System.out.println (root.data ());
 func (root.right ());
}

None of these

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Pre-order Traversal

In this traversal method, the root node is visited first, hence the name. First we traverse the the root node, then left subtree and finally right sub tree.

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Question 10

Time: 00:00:00

What will be the output of the following pseudo code ?

initialize char c

set c = ‘f’

print "%d",c

102

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Since the ASCII value of f is 102 so we get the output as 102.

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