# How to Create Mergable Stack?

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We have to design and create a stack that performs the operations in constant time. Here we have one problem which is how to create mergable stack? Here we perform the below operation for merge two stacks.

• push(element): Insert the element in the stack.
• pop(): Remove the top element in the stack.
• merge (stack 1, stack 2): Merge or join 2 stacks.

## Example

Here is some example of creating a mergable stack please have a look.

Input

push(1, s1);
push(2, s1);
push(3, s1);
push(4, s2);
push(5, s2);
push(6, s2);
merge(s1, s2);
display(s1);

Output

Merged Stack : 3 2 1 6 5 4

Input

push(1, s1);
push(5, s2);
push(6, s2);
merge(s1, s2);
display(s1);

Output

Merged Stack : 1 6 5

## Algorithm

Here we first create two stacks and then try making mergable stack.

1. Create two stacks s1 and s2.
2. Create a function push that accepts integer value and stack as parameter. Initialize a node in it. update data of the new node as given integer and link it to the head of the stack.
3. If the head of the stack is null update tail of stack as a new node. Else update the head of the stack as a new node.
4. Create a function pop that accepts stack as parameter. Check if the head of the stack is null print “stack underflow” else store the head of the stack in a new node, update head as next of head. Return the data of the new node and delete the node.
5. Create a function merge which accepts two stacks as a parameter. Check if the head of stack 1 is null update it’s head as a head of stack 2 and tail as the tail of stack 2 and returns.
6. Else update next of the tail of stack 1 as head of stack 2 and tail of stack 1 as the tail of stack 2.

## C++ Program to create mergable stack

```#include <iostream>
using namespace std;

class node{
public:
int data;
node* next;
};

class newStack{
public:
node* tail;

newStack(){
tail = NULL;
}
};

newStack* create(){
newStack* ns = new newStack();
return ns;
}

void push(int data, newStack* ns){
node* temp = new node();
temp->data = data;

ns->tail = temp;

}

int pop(newStack* ns){
cout << "stack underflow" << endl;
return 0;
}
else{
int popped = temp->data;
delete temp;
return popped;
}
}

void merge(newStack* ns1, newStack* ns2){
ns1->tail = ns2->tail;
return;
}

ns1->tail = ns2->tail;
}

void display(newStack* ns){
cout<<"Merged Stack : ";
while(temp != NULL) {
cout << temp->data << " ";
temp = temp->next;
}
}

int main(){
newStack* s1 = create();
newStack* s2 = create();

push(1, s1);
push(2, s1);
push(3, s1);

push(4, s2);
push(5, s2);
push(6, s2);

merge(s1, s2);
display(s1);

return 0;
}```
`Merged Stack : 3 2 1 6 5 4`

## Complexity Analysis to create mergable stack

Time Complexity: O(1) as all the operations are using constant time i.e. O(1).

Auxiliary Space: O(1) because we are using no extra space.

References

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