Data Structure I Mock Test 2

10%

Question 1 of 10

1. In linked list implementation of a queue, where does a new element be inserted?

At the tail of the link list

At any position in the linked list

At the centre position in the link list

At the head of link list

Question 1 of 10

Question 2 of 10

2. Express -15 as a 6-bit signed binary number.

101110

001111

001110

101111

Question 2 of 10

Question 3 of 10

3. In the given C snippet, find the statement number that has error.

//C code to push an element into a stack
1. void push( struct stack *s, int x) 
2. {
3.     if(s->top==MAX-1)
4.     {
5.         printf(“stack overflow”);
6.     }
7.     else
8.     {
9.         s->items[++s->top]=x;
10.        s++;
11.    }   
12. }

Question 3 of 10

Question 4 of 10

4. Write a piece of code which returns true if the string contains balanced parenthesis, false otherwise.

public boolean isBalanced(String exp)
{
	int len = exp.length();
	Stack<Integer> stk = new Stack<Integer>();
	for(int i = 0; i < len; i++)
        {
		char ch = exp.charAt(i);
                if (ch == '(')
                stk.push(i);
                else if (ch == ')')
                {
			if(stk.peek() != null)
                        {
				return false;
			}
			stk.pop();
		}
	}
	return true;
  }

public boolean isBalanced(String exp)
{
	int len = exp.length();
	Stack<Integer> stk = new Stack<Integer>();
	for(int i = 0; i < len; i++)
        {
	       char ch = exp.charAt(i);
               if (ch == ')')
               stk.push(i);
               else if (ch == '(')
               {
			if(stk.peek() == null)
                        {
				return false;
			}
			stk.pop();
		}
	}
	return true;
}

public boolean isBalanced(String exp)
{
	int len = exp.length();
	Stack<Integer> stk = new Stack<Integer>();
	for(int i = 0; i < len; i++)
        {
		char ch = exp.charAt(i);
                if (ch == '(')
                stk.push(i);
                else if (ch == ')')
                {
			if(stk.peek() == null)
                        {
				return false;
			}
			stk.pop();
		}
	}
	return true;
}

public boolean isBalanced(String exp)
{
	int len = exp.length();
	Stack<Integer> stk = new Stack<Integer>();
	for(int i = 0; i < len; i++)
            {
		char ch = exp.charAt(i);
                if (ch == '(')
                stk.push(i);
                else if (ch == ')')
                {
			if(stk.peek() != null)
                        {
				return true;
			}
			stk.pop();
		}
	}
	return false;
  }

Question 4 of 10

Question 5 of 10

5. Find the output of the following prefix expression.

*+2-2 1/-4 2+-5 3 1

Question 5 of 10

Question 6 of 10

6. What would be the Prefix notation for the given equation?

A^B^C^D

ABCD^^^

^A^B^CD

AB^C^D

^^^ABCD

Question 6 of 10

Question 7 of 10

7. In Postfix expressions, the operators come after the operands.

True

False

Question 7 of 10

Question 8 of 10

8. What is the time complexity of the following code?

public boolean isBalanced(String exp)
{
	int len = exp.length();
	Stack<Integer> stk = new Stack<Integer>();
	for(int i = 0; i < len; i++)
        {
		char ch = exp.charAt(i);
                if (ch == '(')
                stk.push(i);
                else if (ch == ')')
                {
			if(stk.peek() == null)
                        {
				return false;
			}
			stk.pop();
		}
	}
	return true;
}

O(logn)

O(1)

O(n)

O(nlogn)

Question 8 of 10

Question 9 of 10

9. Identify the infix expression from the list of options given below.

ab-c-

abc*+d+ab+cd+*ce-f-

a/b+(c-d)

+ab

Question 9 of 10

Question 10 of 10

10. Making the push operation costly, select the code snippet which implements the pop operation.

public void pop()
{
        if(q2.size()>0)
        {
            q1.poll();
        }
        else
        {
            q2.poll();
        }
}

public void pop()
{
        if(q1.size()>0)
        {
            q1.poll();
        }
        else if(q2.size()>0)
        {
            q2.poll();
        }
}

public void pop()
{
        q1.poll();
	q2.poll();
}

public void pop()
{
        if(q1.size()>0)
        {
            q2.poll();
        }
        else if(q2.size()>0)
        {
            q1.poll();
        }
}

Question 10 of 10

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