Data structure problem 1:Implement a queue using 2 stacks.



This is one of the interview questions asked in technical or data structure round.

As we know stack is first in last out and queue is first in first out.

Here we have to implement the behaviour of queue using 2 stacks.

So let's begin with one of the approaches,


For 2 stacks we have define 2 arrays;


var stack1=[];

var stack2=[];



For insertion or push, we will use stack1.

For any insertion, we can push it in stack1 array.


a)Insert "A"

stack1.push("A");


b)Insert "B"

stack1.push("B");


c)Insert "C"

stack1.push("C");


Now when we remove or pop, the first pushed element should be removed first i.e behaviour of a queue.

  

So if we poped from stack1 we will get the behaviour of stack here.


stack1.pop()//"C" will be removed i.e behaviour of stack last in first out.


We can't use this, so we will need stack2 for this purpose to implement the behaviour of the queue.


d)Remove or pop:

For first remove or pop, we will need to pop every element in stack1 and push it to stack2.

So here are the values of stack2 and stack1, 

   stack2 =["C","B","A"];

   stack1=[];

And then pop from stack2. 

i.e stack2.pop()//"A" will be poped which is inserted first. 

So here we have implemented behaviour of queue.



For another insertion or push,

e)Insert "D"

stack1.push("D");


stack1=["D"];

stack2=["C","B"];


Now for 2nd remove operation first we need to check if stack2 is empty. 

If not empty then just pop element from stack2 for every remove

operation till stack2 becomes empty.

stack2.pop()//=>"B"

stack2.pop()//=>"C"

Now stack 2 is empty, So we can't pop .

stack1=["D"];

stack2=[];


So if stack2 is empty then pushed all elements from stack1 to stack2 and pop stack2.

i.e one that we had done for first remove or pop operation.  

stack1=[];

stack2=["D"];

stack2.pop()//=>"D"


By this approach, we can implement a queue using two stacks.


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Written By:

Saurabh Joshi




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