Compruebe si los elementos de la cola son consecutivos por pares | Conjunto-2

Dada una cola de enteros. La tarea es verificar si los elementos consecutivos en la cola son consecutivos por pares.

Ejemplos: 

Input: 1 2 5 6 9 10
Output: Yes

Input: 2 3 9 11 8 7
Output: No

Acercarse : 

  • Tome una variable n para almacenar el tamaño de la cola.
  • Empuje un elemento a la cola que actúa como marcador.
  • Ahora, si el tamaño de la cola es impar, comience a hacer estallar un par mientras n > 2. Además, mientras hace estallar estos pares, verifique su diferencia y disminuya n en 2, si la diferencia no es igual a 1, establezca el indicador en falso.
  • Si el tamaño de la cola es par, comience a extraer un par mientras n > 1. Además, mientras extrae estos pares, verifique su diferencia y disminuya n en 2, si su diferencia no es igual a 1, establezca el indicador en falso.
  • Finalmente, marque la bandera, si es verdadera, significa que los elementos en la cola están ordenados por pares, de lo contrario no.

A continuación se muestra la implementación del enfoque anterior: 

C++

// C++ program to check if successive
// pair of numbers in the queue are
// consecutive or not
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if elements are
// pairwise consecutive in queue
bool pairWiseConsecutive(queue<int> q)
{
    // Fetching size of queue
    int n = q.size();
 
    // Pushing extra element to the queue
    // which acts as marker
    q.push(INT_MIN);
 
    // Result flag
    bool result = true;
 
    // If number of elements in the
    // queue is odd pop elements while
    // its size is greater than 2.
    // Else, pop elements while its
    // size is greater than 1
    if (n % 2 != 0) {
 
        while (n > 2) {
            int x = q.front();
            q.pop();
            q.push(x);
            n--;
 
            int y = q.front();
            q.pop();
            q.push(y);
            n--;
 
            if (abs(x - y) != 1) {
                result = false;
            }
        }
 
        // Popping the last element of queue
        // and pushing it again.
        // Also, popping the extra element
        // that we have pushed as marker
        q.push(q.front());
        q.pop();
        q.pop();
    }
 
    else {
        while (n > 1) {
            int x = q.front();
            q.pop();
            q.push(x);
            n--;
 
            int y = q.front();
            q.pop();
            q.push(y);
            n--;
 
            if (abs(x - y) != 1) {
                result = false;
            }
        }
 
        // popping the extra element that
        // we have pushed as marker
        q.pop();
    }
 
    return result;
}
 
// Driver program
int main()
{
    // Pushing elements into the queue
    queue<int> q;
    q.push(4);
    q.push(5);
    q.push(-2);
    q.push(-3);
    q.push(11);
    q.push(10);
    q.push(5);
    q.push(6);
    q.push(8);
 
    if (pairWiseConsecutive(q))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
 
    return 0;
}

Java

import java.util.LinkedList;
import java.util.Queue;
 
// Java program to check if successive
// pair of numbers in the queue are
// consecutive or not
public class GFG {
 
// Function to check if elements are
// pairwise consecutive in queue
    static boolean pairWiseConsecutive(Queue<Integer> q) {
        // Fetching size of queue
        int n = q.size();
 
        // Pushing extra element to the queue
        // which acts as marker
        q.add(Integer.MAX_VALUE);
 
        // Result flag
        boolean result = true;
 
        // If number of elements in the
        // queue is odd pop elements while
        // its size is greater than 2.
        // Else, pop elements while its
        // size is greater than 1
        if (n % 2 != 0) {
 
            while (n > 2) {
                int x = q.peek();
                q.remove();
                q.add(x);
                n--;
 
                int y = q.peek();
                q.remove();
                q.add(y);
                n--;
 
                if (Math.abs(x - y) != 1) {
                    result = false;
                }
            }
 
            // Popping the last element of queue
            // and pushing it again.
            // Also, popping the extra element
            // that we have pushed as marker
            q.add(q.peek());
            q.remove();
            q.remove();
        } else {
            while (n > 1) {
                int x = q.peek();
                q.remove();
                q.add(x);
                n--;
 
                int y = q.peek();
                q.remove();
                q.add(y);
                n--;
 
                if (Math.abs(x - y) != 1) {
                    result = false;
                }
            }
 
            // popping the extra element that
            // we have pushed as marker
            q.remove();
        }
 
        return result;
    }
 
// Driver program
    static public void main(String[] args) {
        // Pushing elements into the queue
        Queue<Integer> q = new LinkedList<>();
        q.add(4);
        q.add(5);
        q.add(-2);
        q.add(-3);
        q.add(11);
        q.add(10);
        q.add(5);
        q.add(6);
        q.add(8);
 
        if (pairWiseConsecutive(q)) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
 
    }
}
// This code is contributed by Rajput-Ji

Python3

# Python3 program to check if successive
# pair of numbers in the queue are
# consecutive or not
import sys, queue
 
# Function to check if elements are
# pairwise consecutive in queue
def pairWiseConsecutive(q):
     
    # Fetching size of queue
    n = q.qsize()
 
    # Pushing extra element to the
    # queue which acts as marker
    q.put(-sys.maxsize);
 
    # Result flag
    result = bool(True)
 
    # If number of elements in the
    # queue is odd pop elements while
    # its size is greater than 2.
    # Else, pop elements while its
    # size is greater than 1
    if (n % 2 != 0):
        while (n > 2):
            x = q.queue[0]
            q.get()
            q.put(x)
            n -= 1
 
            y = q.queue[0]
            q.get()
            q.put(y)
            n -= 1
 
            if (abs(x - y) != 1):
                result = bool(False)
 
        # Popping the last element of queue
        # and pushing it again.
        # Also, popping the extra element
        # that we have pushed as marker
        q.put(q.queue[0])
        q.get()
        q.get()
         
    else:
        while (n > 1):
            x = q.queue[0]
            q.get()
            q.put(x)
            n -= 1
 
            y = q.queue[0] 
            q.get()
            q.put(y)
            n -= 1
 
            if (abs(x - y) != 1):
                result = bool(False)
     
        # popping the extra element that
        # we have pushed as marker
        q.get()
 
    return result
 
# Driver code
 
# Pushing elements into the queue
q = queue.Queue()
q.put(4)
q.put(5)
q.put(-2)
q.put(-3)
q.put(11)
q.put(10)
q.put(5)
q.put(6)
q.put(8)
 
if (bool(pairWiseConsecutive(q))):
    print("Yes")
else:
    print("No")
 
# This code is contributed by divyeshrabadiya07

C#

// C# program to check if successive
// pair of numbers in the queue are
// consecutive or not
using System;
using System.Collections.Generic;
 
class GFG
{
 
    // Function to check if elements are
    // pairwise consecutive in queue
    static Boolean pairWiseConsecutive(Queue<int> q)
    {
        // Fetching size of queue
        int n = q.Count;
 
        // Pushing extra element to the queue
        // which acts as marker
        q.Enqueue(int.MaxValue);
 
        // Result flag
        Boolean result = true;
 
        // If number of elements in the
        // queue is odd pop elements while
        // its size is greater than 2.
        // Else, pop elements while its
        // size is greater than 1
        if (n % 2 != 0)
        {
            while (n > 2)
            {
                int x = q.Peek();
                q.Dequeue();
                q.Enqueue(x);
                n--;
 
                int y = q.Peek();
                q.Dequeue();
                q.Enqueue(y);
                n--;
 
                if (Math.Abs(x - y) != 1)
                {
                    result = false;
                }
            }
 
            // Popping the last element of queue
            // and pushing it again.
            // Also, popping the extra element
            // that we have pushed as marker
            q.Enqueue(q.Peek());
            q.Dequeue();
            q.Dequeue();
        } else
        {
            while (n > 1)
            {
                int x = q.Peek();
                q.Dequeue();
                q.Enqueue(x);
                n--;
 
                int y = q.Peek();
                q.Dequeue();
                q.Enqueue(y);
                n--;
 
                if (Math.Abs(x - y) != 1)
                {
                    result = false;
                }
            }
 
            // popping the extra element that
            // we have pushed as marker
            q.Dequeue();
        }
 
        return result;
    }
 
    // Driver Code
    static public void Main(String[] args)
    {
        // Pushing elements into the queue
        Queue<int> q = new Queue<int>();
        q.Enqueue(4);
        q.Enqueue(5);
        q.Enqueue(-2);
        q.Enqueue(-3);
        q.Enqueue(11);
        q.Enqueue(10);
        q.Enqueue(5);
        q.Enqueue(6);
        q.Enqueue(8);
 
        if (pairWiseConsecutive(q))
        {
            Console.WriteLine("Yes");
        }
        else
        {
            Console.WriteLine("No");
        }
    }
}
 
// This code is contributed by Rajput-Ji
Producción: 

Yes

 

Complejidad temporal: O(n) 
Espacio auxiliar: O(1)
 

Publicación traducida automáticamente

Artículo escrito por barykrg y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA

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