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+28 votes

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 $top2$ $(top < top 2)$ 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

  1. $(top1 = MAXSIZE / 2)$ and $(top2 = MAXSIZE / 2 + 1)$

  2. $top1 + top2 = MAXSIZE$

  3. $(top1 = MAXSIZE / 2)$ or $(top2 = MAXSIZE)$

  4. $top1 = top2 - 1$

in DS by Veteran (52.2k points)
edited by | 5.1k views
This question was asked in BARC 2019

I couldn't add a tag to the question, probably because I don't have the question-editing privilege


(Top1 < Top2)

6 Answers

+43 votes
Best answer

Answer is (D).

Since the stacks are growing from opposite ends, initially $top1 = 1$ and $top2 = MAXSIZE$. Now, to make the space usage most efficient we should allow one stack to use the maximum possible space as long as other stack doesn't need it. So, either of the stack can grow as long as there is space on the array and hence the condition must be $top1 = top2 - 1$.

by Loyal (5.2k points)
edited by

@arjun SIR , 

The two stacks grow from opposite ends of the array. 

top 1 grows from begining and top 2 grows from end  the moment it crosses is top 1 <= N-top 2 then Option B should be correct ?


Gokhu top1 is 1st index and top 2 is last index . sum is maxsize is not for full stack condition na.

top1                                              top2

now seek top1+ top2= maxsize but stack contain 2 elements only.

12      13   14    15     16    17   18      19 (top1)     21(top2)   20  

here top1= top2-1 shows stack is full only.


Can you help me to clear below mentioned stack? Am not able to get " how top1 = top2 - 1 " and also what are these values 12,13,14,15,16,17,18,19,21 and 20 ?

@Prashant.Pethe top1= 7, top2=8(considering the array index starts from 0)

so top1=top2-1.

It's an array containing the elements 12,13,14,15,16,17,18,19,21 and 20 and top1 and top2 mark the top of the two stacks respectively where the 1st stack grows from left to right and the second one vice versa.
+19 votes

Consider array with 10 index:

top1                                              top2

Now seek Top1+ Top2= maxsize but stack contain 2 elements only. So Option B is fail.

12      13   14    15     16    17   18      19 (top1)     21(top2)   20  

here top1= top2-1 shows stack is full only. So Option D is correct.

                          top1                          top2

We see here stack is not full so condition Fail. So Option C is fail.

Option A is says both array have same size but there may be case that stack 2 contain 1 element and stack 2 contain 9 elements , which is not possible with case A.

D is answer

by Veteran (63k points)

@gokou  check here


@anirudh SIR , 

top1                                              top2

Here , How  top1 + top2 = MAX SIZE ? top1 and top2 are stack top they are incremented when new elemets are added to stack  

In one element  stack given above top1=top2= 1  , top1 + top2 = 2 only . 

And here ,

12      13   14    15     16    17   18      19 (top1)     21(top2)   20  

N= 10 , Top1=8 Top2=2
Top1 <= MAX-Top1 holds good . So option B should be true ? 

Please correct me if wrong  


@anirudh SIR , 

Top1 and Top2 is 

top1 and top 2 (top < top 2) point to the location of the topmost element in each of the stacks

But I considered Top1 and Top2 as number of elements in the stacks , is that the reason I arrived at option B ?

I think,option A says both the stack have same size.
+5 votes

The idea is to start two stacks from two extreme corners of arr[]. stack1 starts from the leftmost element, the first element in stack1 is pushed at index 0. The stack2 starts from the rightmost corner, the first element in stack2 is pushed at index (n-1). Both stacks grow (or shrink) in opposite direction. To check for overflow, all we need to check is for space between top elements of both stacks


by Loyal (7.8k points)
0 votes

Look at how heap and stack sections are incorporated in the Runtime Environment.

That is the most efficient way to utilize the space. If we divide the space by half, what if one entity requires 60% of the space, while the other entity requires none? Halving the available space is an inefficient way to utilize it.

Efficient utilization would be to let a stack take up all the space if the other stack doesn't want it.


For the array [1...MaxSize] Let top1 point to 0, and top2 point to MaxSize+1 in the beginning, since both the stacks are empty.

Stack full will be when:-

Option D

by Loyal (6.4k points)
–3 votes
I think both A and D are correct.actually both defining same thing.
by Active (2.9k points)
no Option D is variable in nature but A is fixed to mid point and adjacent to mid.
got it. Thank you.

So D is the answer
@anirudh SIR , Why not B?
Sir, is the midpoint thing not efficient? What is the need for flexibility here?
–4 votes
A ) top1= ⎣MAXSIZE/2⎦

     top2= ⎣MAXSIZE/2⎦+1
by Active (2.6k points)
That would limit the maximum size of each stack to be half of the array.
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