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+38 votes
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An implementation of a queue $Q$, using two stacks $S1$ and $S2$, is given below: 

void insert (Q, x) { 
    push (S1, x); 
} 
void delete (Q) { 
    if (stack-empty(S2)) then 
        if (stack-empty(S1)) then { 
            print(“Q is empty”); 
            return; 
        } 
        else while (!(stack-empty(S1))){ 
            x=pop(S1); 
            push(S2,x); 
        } 
    x=pop(S2);
}

let $n$ insert and $ m(\leq n)$ delete operations be performed in an arbitrary order on an empty queue $Q$. Let $x$ and $y$  be the number of push and pop operations  performed respectively in the process. Which one of the following is true for all $m$ and $n$?

  1. $ n+m\leq x<2n $ and $2m\leq y\leq n+m $
  2. $ n+m\leq x<2n $ and $2m\leq y\leq 2n $
  3. $ 2m\leq x<2n $ and $2m\leq y\leq n+m $
  4. $ 2m\leq x<2n $ and $2m\leq y\leq 2n $
asked in DS by Active (3.7k points)
edited by | 4.3k views
+1
Can someone tell why bcd are not true?Counter examples?
0

@  rahul sharma 5, A is more stronger option than B, i.e it is superset of option B. Options C & D are false because, It is mentioned that there are n insert operations due to which n < x condition should hold always.

0

How did u prove c and d false?I did not get.I found all are true but a being the strongest answer as it gives exact answer but other are upper and lower bounds

0
What will happen if i try to delete and my stack S2 is not empty? Firstly S2 elements should be returned and then if S2 is empty,then S1 should be brought into S2
0

I don't know whether anyone noticed or not,

but for all options for x,

the inequality for X, must have $\leq$ sign on the right part of inequality

that means x must be x$\leq$ 2n instead of x$<$2n

Since question says, for any arbitrary order of insert and delete operations,

consider below case

n=4(the total insertions)

m=3(the total deletions)

Take queue operations as

2 insert, 1 delete, 1 insert , 1 delete , 1 insert, 1 delete,

then x(total number of push) =8

y(total number of pop) = 7

now 2n=8 and x$\nless$8

Question clearly says

Let n insert and m(≤n)m(≤n) delete operations be performed in an arbitrary order on an empty queue Q. Let x and y  be the number of push and pop operations  performed respectively in the process. Which one of the following is true for all m and n?

I think options must be changed considering the inequality.

Let me know if anyone agrees.

0
there is a fault in above program ,

 before inserting anything to stack S1, program should check whether the stack s2 is empty or not , if S2 is  not empty then it should first copy all contents of S2  to S1,  otherwise  queue can not be implemented .

and second thing is that x can be equal to 2n

5 Answers

+40 votes
Best answer

Answer is (A).

The order in which insert and delete operations are performed matters here.

The best case: Insert and delete operations are performed alternatively. In every delete operation, $2$ pop and $1$ push operations are performed. So, total $m+ n$ push $(n$ push for insert() and $m$ push for delete()$)$ operations and $2m$ pop operations are performed.

The worst case: First $n$ elements are inserted and then m elements are deleted. In first delete operation, $n + 1$ pop operations and $n$ push operation are performed. Other than first, in all delete operations, $1$ pop operation is performed. So, total $m + n$ pop operations and $2n$ push operations are performed $(n$ push for insert() and $m$ push for delete()$)$

answered by Active (3.3k points)
edited by
+8
I think x <= 2n instead of x < 2n. Am I correct?
0
Ya, I also think the same!
0
Yes please make the correction or else none of answers are correct. Please make corrections in gateoverflow pdf as well.
+32 votes

Page 1

Page 2

[Please excuse for the poor handwriting ]

answered by Boss (22.1k points)
edited by
0
this is so perfect (Y)
0
can't be better than this
0
Nice one
0
In the second case, why is the overall requirement of pop = 4 ?
0
Why b is not true?

If second condition

 2m≤y≤n+m is true then

 2m≤y≤2n will also be true as RHS cannot be more than 2n >=n+m
0
@rahul But y can never be equal to 2n. So y≤2n is not correct.
0
In page 2

what do you mean by for m elements to come out = 2pushes

remaining elements = n-m
+16 votes
the order in which we insert and delete is what matters here.We will have to find the best and the worst cases possible:--

1.WORST CASE-- The worst case happens when we first insert n elements into say STACK 1. Now this n insertions needs n PUSH operations on STACK 1.Now to delete m elements from this n elements we will have to first POP out this n elements from STACK 1.This takes n POP operations.NOW, again we have to PUSH these n elements in STACK 2 in reverse order(as QUEUE follows FIFO property).thus because of this we get n more PUSH operations in STACK 2.Now,finally we POP m elements from STACK 2 to delete m elements.Thus, total PUSH operations becomes 2n and total POP operations becomes n+m.

2.BEST CASE--The best case occurs when we first insert those elements which are to be deleted in STACK 1.Thus, this takes m PUSH operations.Now to delete these m elements we have to first POP these from STACK 1, which takes m POP operations. Now PUSH these m elements in STACK 2 in reverse order as before.This takes m PUSH operations more.Now again POP these m elements to delete it.But now in the question it says that the number of insertions should be n. But we have only inserted m of it.SO, now insert more n-m elements in STACK 1. Thus the total PUSH operations becomes m+m+n-m=n+m and the total POP operations becomes 2m.

Thus, option A is the answer..
answered by Active (4k points)
0
I understood the first part but not the second one , so can u clarify it again please why did u consider initially m push operations and m pop operations when we have n insertions to take place.
0
so that the best case occurs..you may check that if the order of insertion and deletion is in this sequence then the best case occurs...
+5 votes

Elements to be pushed = $n$
Elements to be popped = $m$

$m < n$
 x = number of PUSH operations
 y = number of POP operations

case 1: range of y 
lower bound:
1. push m elements to stack S1
2. Pop m elements from stack s1.
3. push m elements on stack s2.
4. pop all m elements from stack s2.

Total pop operations happened here are $m+m = 2m$, hence $2m\leq y$

Upper Bound:
1. Push all n elements onto stack s1.
2. pop all n elements from stack s1.
3. push all elements on stack s2.
4. pop m elements from stack s2.

Total pop operations happened here are n+m, hence $y\leq m+n$

So far so good to eliminate the options (B) and (D).

Range of x:

lower bound:
1. push m elements to stack S1
2. Pop m elements from stack s1.
3. push m elements on stack s2.
4. pop all m elements from stack s2.
5. push remaining n-m elements on stack s2.

total push operations = $m + m + n - m = n+m$, hence $m+n \leq x$

So far so good to eliminate the option (C).

Upper Bound:
1. Push all n elements onto stack s1.
2. pop all n elements from stack s1.
3. push all n elements on stack s2.
4. pop m elements from stack s2.

Total PUSH operations = n+n, hence $x \leq 2n$ 

Hence, (A) is the correct choice.

PS: Don't know why is equal operator not given in any option.
 

answered by Boss (39.8k points)
+4 votes
void insert (Q, x) { 
    push (S1, x); 
} 
void delete (Q) { 
    if (stack-empty(S2)) then 
        if (stack-empty(S1)) then { 
            print(“Q is empty”); 
            return; 
        } 
        else while (!(stack-empty(S1))){ 
            x=pop(S1); 
            push(S2,x); 
        } 
    x=pop(S2);
}

See PUSH  and POP operation in code.

Given : n=insert

m=delete

x=push

y=pop

Trywith some example.

Now, in given code 

Say S_2 already contains 1,2,3 // No PUSH operation here, no need to consider it

$S_{1}$ has to push 4,5 // So, 2 PUSH operation here.

Now, POP S_2 //3 POP operation Not considering here

Now, POP $S_1$// 2 POP operation

PUSH them in $S_2$//2 PUSH operation

Again POP $S_2$// 2 POP operation

-------------------------------------------------------------------------------------------------------------------------------

In PUSH 4,5 push to insert $S_1$ (Say, n push)

Now, we only POP 5 from stack (Say m) and PUSH in $S_2$.

So, Total PUSH n+m.

and maximum we can push 2n (when we POP 4,5 both from stack (Say n) and PUSH in $S_2$. )

So, for PUSH operation $n+m< x\leq 2n$

--------------------------------------------------------------------------------------------------------------------------------------------------------------

In POP operation say  5  POP from $S_1$.(say m)

Now,  5 PUSH in $S_2$.(Say m)

Now, we have to POP only 5 from $S_2$.

Otherwise in worst case

In POP operation say  4,5  POP from $S_1$.(say n)

Now, only 5 PUSH in $S_2$.(Say m)

Now, we have to POP only 5 from $S_2$.

n+m POP operation

So, total POP operation $2m< x\leq m+n$

answered by Veteran (91.7k points)
edited by


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