Question 1 
Following is C like pseudo code of a function that takes a number as an argument, and uses a stack S to do processing.
void fun(int n) { Stack S; // Say it creates an empty stack S while (n > 0) { // This line pushes the value of n%2 to stack S push(&S, n%2); n = n/2; } // Run while Stack S is not empty while (!isEmpty(&S)) printf("%d ", pop(&S)); // pop an element from S and print it }What does the above function do in general?
Prints binary representation of n in reverse order  
Prints binary representation of n  
Prints the value of Logn  
Prints the value of Logn in reverse order 
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Question 1 Explanation:
See method 2 of http://www.geeksforgeeks.org/binaryrepresentationofagivennumber/ for explanation.
Question 2 
Which one of the following is an application of Stack Data Structure?
Managing function calls  
The stock span problem  
Arithmetic expression evaluation  
All of the above 
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Question 2 Explanation:
Question 3 
Which of the following is true about linked list implementation of stack?
In push operation, if new nodes are inserted at the beginning of linked list, then in pop operation, nodes must be removed from end.  
In push operation, if new nodes are inserted at the end, then in pop operation, nodes must be removed from the beginning.  
Both of the above  
None of the above 
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Question 3 Explanation:
To keep the Last In First Out order, a stack can be implemented using linked list in two ways:
a) In push operation, if new nodes are inserted at the beginning of linked list, then in pop operation, nodes must be removed from beginning.
b) In push operation, if new nodes are inserted at the end of linked list, then in pop operation, nodes must be removed from end.
Question 4 
Consider the following pseudocode that uses a stack
declare a stack of characters while ( there are more characters in the word to read ) { read a character push the character on the stack } while ( the stack is not empty ) { pop a character off the stack write the character to the screen }What is output for input "geeksquiz"?
geeksquizgeeksquiz  
ziuqskeeg  
geeksquiz  
ziuqskeegziuqskeeg 
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Question 4 Explanation:
Since the stack data structure follows LIFO order. When we pop() items from stack, they are popped in reverse order of their insertion (or push())
Question 5 
Following is an incorrect pseudocode for the algorithm which is supposed to determine whether a sequence of parentheses is balanced:
declare a character stack while ( more input is available) { read a character if ( the character is a '(' ) push it on the stack else if ( the character is a ')' and the stack is not empty ) pop a character off the stack else print "unbalanced" and exit } print "balanced"Which of these unbalanced sequences does the above code think is balanced? Source: http://www.cs.colorado.edu/~main/questions/chap07q.html
((())  
())(()  
(()()))  
(()))() 
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Question 5 Explanation:
At the end of while loop, we must check whether the stack is empty or not. For input ((()), the stack doesn't remain empty after the loop. See http://www.geeksforgeeks.org/checkforbalancedparenthesesinanexpression/ for details.
Question 6 
The following postfix expression with single digit operands is evaluated using a stack:
8 2 3 ^ / 2 3 * + 5 1 * Note that ^ is the exponentiation operator. The top two elements of the stack after the first * is evaluated are:
6, 1  
5, 7  
3, 2  
1, 5 
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Question 6 Explanation:
The algorithm for evaluating any postfix expression is fairly straightforward:
1. While there are input tokens left o Read the next token from input. o If the token is a value + Push it onto the stack. o Otherwise, the token is an operator (operator here includes both operators, and functions). * It is known a priori that the operator takes n arguments. * If there are fewer than n values on the stack (Error) The user has not input sufficient values in the expression. * Else, Pop the top n values from the stack. * Evaluate the operator, with the values as arguments. * Push the returned results, if any, back onto the stack. 2. If there is only one value in the stack o That value is the result of the calculation. 3. If there are more values in the stack o (Error) The user input has too many values.Source for algorithm: http://en.wikipedia.org/wiki/Reverse_Polish_notation#The_postfix_algorithm Let us run the above algorithm for the given expression. First three tokens are values, so they are simply pushed. After pushing 8, 2 and 3, the stack is as follows
8, 2, 3When ^ is read, top two are popped and power(2^3) is calculated
8, 8When / is read, top two are popped and division(8/8) is performed
1Next two tokens are values, so they are simply pushed. After pushing 2 and 3, the stack is as follows
1, 2, 3When * comes, top two are popped and multiplication is performed.
1, 6
Question 7 
Let S be a stack of size n >= 1. Starting with the empty stack, suppose we push the first n natural numbers in sequence, and then perform n pop operations. Assume that Push and Pop operation take X seconds each, and Y seconds elapse between the end of one such stack operation and the start of the next operation. For m >= 1, define the stacklife of m as the time elapsed from the end of Push(m) to the start of the pop operation that removes m from S. The average stacklife of an element of this stack is
n(X+ Y)  
3Y + 2X  
n(X + Y)X  
Y + 2X 
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Question 7 Explanation:
We can easily arrive at the result by taking few examples.
Question 8 
A single array A[1..MAXSIZE] is used to implement two stacks. The two stacks grow from opposite ends of the array. Variables top1 and top2 (topl< top 2) point to the location of the topmost element in each of the stacks. If the space is to be used efficiently, the condition for “stack full” is (GATE CS 2004)
(top1 = MAXSIZE/2) and (top2 = MAXSIZE/2+1)  
top1 + top2 = MAXSIZE  
(top1= MAXSIZE/2) or (top2 = MAXSIZE)  
top1= top2 1 
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Question 8 Explanation:
If we are to use space efficiently then size of the any stack can be more than MAXSIZE/2.
Both stacks will grow from both ends and if any of the stack top reaches near to the other top then stacks are full. So the condition will be top1 = top2 1 (given that top1 < top2)
Question 9 
Assume that the operators +, , × are left associative and ^ is right associative. The order of precedence (from highest to lowest) is ^, x , +, . The postfix expression corresponding to the infix expression a + b × c  d ^ e ^ f is
abc × + def ^ ^   
abc × + de ^ f ^   
ab + c × d  e ^ f ^  
 + a × bc ^ ^ def 
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Question 9 Explanation:
^ is right assosciative.
See http://geeksquiz.com/stackset2infixtopostfix/
This solution is contributed by parul Sharma.
Question 10 
To evaluate an expression without any embedded function calls:
One stack is enough  
Two stacks are needed  
As many stacks as the height of the expression tree are needed  
A Turing machine is needed in the general case 
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Question 10 Explanation:
Any expression can be converted into Postfix or Prefix form.
Prefix and postfix evaluation can be done using a single stack.
For example : Expression '10 2 8 * + 3 ' is given. PUSH 10 in the stack. PUSH 2 in the stack. PUSH 8 in the stack. When operator '*' occurs, POP 2 and 8 from the stack. PUSH 2 * 8 = 16 in the stack. When operator '+' occurs, POP 16 and 10 from the stack. PUSH 10 * 16 = 26 in the stack. PUSH 3 in the stack. When operator '' occurs, POP 26 and 3 from the stack. PUSH 26  3 = 23 in the stack. So, 23 is the answer obtained using single stack.
Thus, option (A) is correct.
Please comment below if you find anything wrong in the above post.
Question 11 
The result evaluating the postfix expression 10 5 + 60 6 / * 8 – is
284  
213  
142  
71 
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Question 11 Explanation:
Question 12 
A function f defined on stacks of integers satisfies the following properties. f(∅) = 0 and f (push (S, i)) = max (f(S), 0) + i for all stacks S and integers i.
If a stack S contains the integers 2, 3, 2, 1, 2 in order from bottom to top, what is f(S)?6  
4  
3  
2 
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Question 12 Explanation:
f(S) = 0, max(f(S), 0) = 0, i = 2 f(S)_{new} = max(f(S), 0) + i = 0 + 2 = 2
f(S) = 2, max(f(S), 0) = 2, i = 3 f(S)_{new} = max(f(S), 0) + i = 2  3 = 1
f(S) = 1, max(f(S), 0) = 0, i = 2 f(S)_{new} = max(f(S), 0) + i = 0 + 2 = 2
f(S) = 2, max(f(S), 0) = 2, i = 1 f(S)_{new} = max(f(S), 0) + i = 2  1 = 1
f(S) = 1, max(f(S), 0) = 1, i = 2 f(S)_{new} = max(f(S), 0) + i = 1 + 2 = 3
Thus, option (C) is correct.
Please comment below if you find anything wrong in the above post.
Question 13 
Consider the following C program:
#include #define EOF 1 void push (int); /* push the argument on the stack */ int pop (void); /* pop the top of the stack */ void flagError (); int main () { int c, m, n, r; while ((c = getchar ()) != EOF) { if (isdigit (c) ) push (c); else if ((c == '+')  (c == '*')) { m = pop (); n = pop (); r = (c == '+') ? n + m : n*m; push (r); } else if (c != ' ') flagError (); } printf("% c", pop ()); }What is the output of the program for the following input ? 5 2 * 3 3 2 + * +
15  
25  
30  
150 
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Question 13 Explanation:
The function of the program is:
1) If the current character is a digit it pushes into stack
2) Else if the current character is operator, it pops two elements and then performs the operation.
Finally it pushes the resultant element into stack.
Initially stack s is empty. 5 2 * 3 3 2 + * +
1) 5 > It pushes into s
2) 2 > It pushes into s
3) * > It pops two elements n = 2, m=5 n*m = 10 It pushes 10 into s
4) 3 > It pushes into s
5) 3 > It pushes into s
6) 2 > It pushes into s
7) + > n=2, m=3 n+m=5 It pushes 5 into s
8) * > n=5, m=3 n*m=15 It pushes 15 into s
9) + > n=15, m=10 n+m = 25 It pushes 25 into s.
Finally the result value is the only element present in stack.
This solution is contributed by Anil Saikrishna Devarasetty.
Result = 25
Question 14 
Suppose a stack is to be implemented with a linked list instead of an array. What would be the effect on the time complexity of the push and pop operations of the stack implemented using linked list (Assuming stack is implemented efficiently)?
O(1) for insertion and O(n) for deletion  
O(1) for insertion and O(1) for deletion
 
O(n) for insertion and O(1) for deletion
 
O(n) for insertion and O(n) for deletion

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Question 14 Explanation:
Stack can be implemented using link list having O(1) bounds for both insertion as well as deletion by inserting and deleting the element from the beginning of the list.
Question 15 
Consider n elements that are equally distributed in k stacks. In each stack, elements of it are arranged in ascending order (min is at the top in each of the stack and then increasing downwards).
Given a queue of size n in which we have to put all n elements in increasing order. What will be the time complexity of the best known algorithm?
O(n logk)  
O(nk)  
O(n^{2})  
O(k^{2}) 
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Question 15 Explanation:
In nlogk it can be done by creating a min heap of size k and adding all the top  elements of all the stacks. After extracting the min , add the next element from the stack from which we have got our 1st minimum.
Time Complexity = O(k) (For Creating Heap of size k) + (nk)log k (Insertions into the heap).
There are 15 questions to complete.