HackerRank Queues: A Tale of Two Stacks solution

In this HackerRank Queues: A Tale of Two Stacks Interview preparation kit problem you must first implement a queue using two stacks.

Problem solution in Python programming.

class MyQueue(object):
    def __init__(self):
        self.items=[]
    
    def peek(self):
        i = self.items.pop()
        self.items.append(i)
        return i
        
    def pop(self):
        return self.items.pop()
        
    def put(self, value):
        self.items.insert(0, value)
        

queue = MyQueue()
t = int(input())
for line in range(t):
    values = map(int, input().split())
    values = list(values)
    if values[0] == 1:
        queue.put(values[1])        
    elif values[0] == 2:
        queue.pop()
    else:
        print(queue.peek())
        

Problem solution in Java Programming.

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) {
        MyQueue<Integer> queue = new MyQueue<Integer>();

        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();

        for (int i = 0; i < n; i++) {
            int operation = scan.nextInt();
            if (operation == 1) { // enqueue
              queue.enqueue(scan.nextInt());
            } else if (operation == 2) { // dequeue
              queue.dequeue();
            } else if (operation == 3) { // print/peek
              System.out.println(queue.peek());
            }
        }
        scan.close();
    }
    
    static class MyQueue<T> {
        private Stack<T> reversed;
        private Stack<T> normal;

        public MyQueue() {
            normal = new Stack<>();
            reversed = new Stack<>();
        }

        public void enqueue(T item) {
            reversed.push(item);
        }

        private void pour() {
            while (!reversed.isEmpty()) {
                normal.push(reversed.pop());
            }
        }

        public T peek() {
            if (normal.isEmpty()) { pour(); }
            return normal.peek();
        }

        public T dequeue() {
            if (normal.isEmpty()) { pour(); }
            return normal.pop();
        }
    }

}

Problem solution in C++ programming.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include <stack>
#include <queue>
using namespace std;

class MyQueue {
    public:
        stack<int> stack_newest_on_top, stack_oldest_on_top;   
        void push(int x) {
            stack_newest_on_top.push(x);
        }

        void pop() {
            prepOld();
            stack_oldest_on_top.pop();
            prepNew();
        }

        int front() {
            prepOld();
            return stack_oldest_on_top.top();
            prepNew();
        }

        void prepOld(){
            if (stack_oldest_on_top.empty())
            {   
                while(!stack_newest_on_top.empty())
                {
                    stack_oldest_on_top.push(stack_newest_on_top.top());
                    stack_newest_on_top.pop();
                }
            }
        }
        void prepNew(){
            if (stack_newest_on_top.empty())
            {   
                while(!stack_oldest_on_top.empty())
                {
                    stack_newest_on_top.push(stack_oldest_on_top.top());
                    stack_oldest_on_top.pop();
                }
            }
        }

};

int main() {
    MyQueue q1;
    int q, type, x;
    cin >> q;
    
    for(int i = 0; i < q; i++) {
        cin >> type;
        if(type == 1) {
            cin >> x;
            q1.push(x);
        }
        else if(type == 2) {
            q1.pop();
        }
        else cout << q1.front() << endl;
    }
    /* Enter your code here. Read input from STDIN. Print output to STDOUT */  

    return 0;
}

Problem solution in C programming.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

#define MAX 100000

typedef struct {
    int first;
    int last;
    int data[MAX];
} Q;

void Init(Q* q) {
    q->first = 0;
    q->last  = MAX - 1;
}

void Enqueue(Q* q, int val) {
    q->last = (q->last + 1) % MAX;
    q->data[q->last] = val;
}

int Dqueue(Q* q) {
    int val = q->data[q->first];
    q->first = (q->first + 1) % MAX;
    return val;
}

void PrintHead(Q* q) {
    printf("%dn", q->data[q->first]);
}

int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    Q q;
    Init(&q);
    int n;
    scanf("%d", &n);
    
    for(int i = 0; i < n; i++){
        int op, val;
        scanf("%d", &op);
        switch(op) {
            case 1:
                scanf("%d", &val);
                Enqueue(&q, val);
                break;
            case 2:
                Dqueue(&q);
                break;
            case 3:
                PrintHead(&q);
                break;
        }
    }
    return 0;
}

Problem solution in JavaScript programming.

function processData(input) {
    let inbox = [];
    let outbox = [];
    let operations = input.split('n').slice(1).map(line => { return line.split(' ') });
    for (let i = 0; i < operations.length; i++) {
        if (operations[i].length == 2) {
            inbox.push(operations[i][1]);
        } else {
            if (outbox.length === 0) {
                while (inbox.length > 0) {
                    outbox.push(inbox.pop());
                }
            }
            if (operations[i][0] === '2') {
                outbox.pop();
            } else {
                console.log(outbox[outbox.length - 1]);
            }
        }
    }
} 

process.stdin.resume();
process.stdin.setEncoding("ascii");
_input = "";
process.stdin.on("data", function (input) {
    _input += input;
});

process.stdin.on("end", function () {
   processData(_input);
});