# Queue using Two Stacks

### Problem Statement :

```A queue is an abstract data type that maintains the order in which elements were added to it, allowing the oldest elements to be removed from the front and new elements to be added to the rear. This is called a First-In-First-Out (FIFO) data structure because the first element added to the queue (i.e., the one that has been waiting the longest) is always the first one to be removed.

A basic queue has the following operations:

Enqueue: add a new element to the end of the queue.
Dequeue: remove the element from the front of the queue and return it.
In this challenge, you must first implement a queue using two stacks. Then process q queries, where each query is one of the following 3  types:

1 x: Enqueue element x into the end of the queue.
2: Dequeue the element at the front of the queue.
3: Print the element at the front of the queue.
Input Format

The first line contains a single integer, q, denoting the number of queries.
Each line i of the q subsequent lines contains a single query in the form described in the problem statement above. All three queries start with an integer denoting the query type, but only query 1 is followed by an additional space-separated value, x , denoting the value to be enqueued.

Output Format

For each query of type 3, print the value of the element at the front of the queue on a new line.```

### Solution :

```                            ```Solution in C :

In C ++ :

#include <iostream>
#include <stack>

using namespace std;

int main()
{
int q, type, value;
char operation;
stack<int> S1, S2;
stack<char> S3, S4;

cin >> q;
while(q--)
{
cin >> type;
if(type == 1)
{
cin >> value;
S1.push(value);
}
else if(type == 2)
{
S3.push('d');
}
else
{
S3.push('p');
}
}

// Perform the actual operations.
while(!S1.empty())
{
value = S1.top();
S1.pop();
S2.push(value);
}

while(!S3.empty())
{
value = S3.top();
S3.pop();
S4.push(value);
}

while(!S2.empty())
{
operation = S4.top();
value = S2.top();
S4.pop();

if(operation == 'p')
{
// Print.
cout << value << endl;
}
else
{
// Dequeue.
S2.pop();
}
}

return 0;
}

In Java :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int q = in.nextInt();
while (q-- > 0) {
int type = in.nextInt();
if (type == 1) {
}
else if (type == 2) {
if (!l.isEmpty()) {
l.pollFirst();
}
}
else {
System.out.println(l.getFirst());
}
}
}
}

In C :

#include<stdio.h>
#include<malloc.h>

struct node{
int data;
struct node* next;
};

struct queue{
};

struct node* push(struct node* temp,int data){
struct node* newnode = (struct node*)malloc(sizeof(struct node));
newnode->data = data;
newnode->next = temp;
temp = newnode;
return temp;
}

void enqueue(struct queue* q,int data){
}

int pop(struct node** temp){
int data = ( *temp )->data;
(*temp) = (*temp)->next;
return data;
}

void dequeue(struct queue*q){
int x;
}
}else{
}
}

void print(struct queue* q){
int x;
}else{
}

}
printf("%d\n",x);
}

int main(void){
int data,queries,choice;
struct queue* q  = (struct queue*)malloc(sizeof(struct queue));
scanf("%d",&queries);
for(int i = 0 ;i < queries ;i++){
scanf("%d",&choice);
if(choice == 1){
scanf("%d",&data);
enqueue(q,data);
}else if(choice == 2){
dequeue(q);
}else if(choice == 3){
print(q);
}
}

}

In Python3 :

q=int(input())
qa=[]
for q0 in range(q):
li=[int(i) for i in input().strip().split()]
if li==1:
qa.append(li)
elif li==2:
del qa
else:
print(qa)```
```

## View More Similar Problems

Given a pointer to the head of a linked list, insert a new node before the head. The next value in the new node should point to head and the data value should be replaced with a given value. Return a reference to the new head of the list. The head pointer given may be null meaning that the initial list is empty. Function Description: Complete the function insertNodeAtHead in the editor below

## Insert a node at a specific position in a linked list

Given the pointer to the head node of a linked list and an integer to insert at a certain position, create a new node with the given integer as its data attribute, insert this node at the desired position and return the head node. A position of 0 indicates head, a position of 1 indicates one node away from the head and so on. The head pointer given may be null meaning that the initial list is e

## Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

## Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing