Give an algorithm for reversing a queue Q. Only the following standard operations are allowed on queue.
- enqueue(x): Add an item x to the rear of the queue.
- dequeue(): Remove an item from the front of the queue.
- empty(): Checks if a queue is empty or not.
The task is to reverse the queue.
Examples:
Input: Q = [10, 20, 30, 40, 50, 60, 70, 80, 90, 100]
Output: Q = [100, 90, 80, 70, 60, 50, 40, 30, 20, 10]
Input: [1, 2, 3, 4, 5]
Output: [5, 4, 3, 2, 1]
Reversing a Queue using stack:
For reversing the queue one approach could be to store the elements of the queue in a temporary data structure in a manner such that if we re-insert the elements in the queue they would get inserted in reverse order. So now our task is to choose such a data structure that can serve the purpose. According to the approach, the data structure should have the property of ‘LIFO’ as the last element to be inserted in the data structure should actually be the first element of the reversed queue.
Follow the below steps to implement the idea:
- Pop the elements from the queue and insert into the stack now topmost element of the stack is the last element of the queue.
- Pop the elements of the stack to insert back into the queue the last element is the first one to be inserted into the queue.
Below is the implementation of above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void Print(queue<int>& Queue)
{
while (!Queue.empty()) {
cout << Queue.front() << " ";
Queue.pop();
}
}
void reverseQueue(queue<int>& Queue)
{
stack<int> Stack;
while (!Queue.empty()) {
Stack.push(Queue.front());
Queue.pop();
}
while (!Stack.empty()) {
Queue.push(Stack.top());
Stack.pop();
}
}
int main()
{
queue<int> Queue;
Queue.push(10);
Queue.push(20);
Queue.push(30);
Queue.push(40);
Queue.push(50);
Queue.push(60);
Queue.push(70);
Queue.push(80);
Queue.push(90);
Queue.push(100);
reverseQueue(Queue);
Print(Queue);
}
|
Java
import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;
public class Queue_reverse {
static Queue<Integer> queue;
static void Print()
{
while (!queue.isEmpty()) {
System.out.print(queue.peek() + ", ");
queue.remove();
}
}
static void reversequeue()
{
Stack<Integer> stack = new Stack<>();
while (!queue.isEmpty()) {
stack.add(queue.peek());
queue.remove();
}
while (!stack.isEmpty()) {
queue.add(stack.peek());
stack.pop();
}
}
public static void main(String args[])
{
queue = new LinkedList<Integer>();
queue.add(10);
queue.add(20);
queue.add(30);
queue.add(40);
queue.add(50);
queue.add(60);
queue.add(70);
queue.add(80);
queue.add(90);
queue.add(100);
reversequeue();
Print();
}
}
|
Python3
from collections import deque
def reversequeue(queue):
Stack = []
while (queue):
Stack.append(queue[0])
queue.popleft()
while (len(Stack) != 0):
queue.append(Stack[-1])
Stack.pop()
if __name__ == '__main__':
queue = deque([10, 20, 30, 40, 50, 60, 70, 80, 90, 100])
reversequeue(queue)
print(queue)
|
C#
using System;
using System.Collections.Generic;
public class GFG {
public static LinkedList<int> queue;
public static void Print()
{
while (queue.Count > 0) {
Console.Write(queue.First.Value + ", ");
queue.RemoveFirst();
}
}
public static void reversequeue()
{
Stack<int> stack = new Stack<int>();
while (queue.Count > 0) {
stack.Push(queue.First.Value);
queue.RemoveFirst();
}
while (stack.Count > 0) {
queue.AddLast(stack.Peek());
stack.Pop();
}
}
public static void Main(string[] args)
{
queue = new LinkedList<int>();
queue.AddLast(10);
queue.AddLast(20);
queue.AddLast(30);
queue.AddLast(40);
queue.AddLast(50);
queue.AddLast(60);
queue.AddLast(70);
queue.AddLast(80);
queue.AddLast(90);
queue.AddLast(100);
reversequeue();
Print();
}
}
|
Javascript
<script>
let queue = [];
function Print()
{
while (queue.length > 0) {
document.write( queue[0] + ", ");
queue.shift();
}
}
function reversequeue()
{
let stack = [];
while (queue.length > 0) {
stack.push(queue[0]);
queue.shift();
}
while (stack.length > 0) {
queue.push(stack[stack.length - 1]);
stack.pop();
}
}
queue = []
queue.push(10);
queue.push(20);
queue.push(30);
queue.push(40);
queue.push(50);
queue.push(60);
queue.push(70);
queue.push(80);
queue.push(90);
queue.push(100);
reversequeue();
Print();
</script>
|
Output100 90 80 70 60 50 40 30 20 10
Time Complexity: O(N), As we need to insert all the elements in the stack and later to the queue.
Auxiliary Space: O(N), Use of stack to store values.
Reversing a Queue using recursion:
Instead of explicitly using stack goal can be achieved using recursion (recursion at backend will itself maintain stack).
Follow the below steps to implement the idea:
- Recursively perform the following steps:
- If the queue size is 0 return.
- Else pop and store the front element and recur for remaining queue.
- push the current element in the queue.
Thank you Nakshatra Chhillar for suggesting this approach and contributing the code
Below is the implementation of above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void Print(queue<int>& Queue)
{
while (!Queue.empty()) {
cout << Queue.front() << " ";
Queue.pop();
}
}
void reverseQueue(queue<int>& q)
{
if (q.size() == 0)
return;
int fr = q.front();
q.pop();
reverseQueue(q);
q.push(fr);
}
int main()
{
queue<int> Queue;
Queue.push(10);
Queue.push(20);
Queue.push(30);
Queue.push(40);
Queue.push(50);
Queue.push(60);
Queue.push(70);
Queue.push(80);
Queue.push(90);
Queue.push(100);
reverseQueue(Queue);
Print(Queue);
}
|
Java
import java.io.*;
import java.util.*;
class GFG {
public static void Print(Queue<Integer> Queue)
{
while (Queue.size() > 0) {
System.out.print(Queue.peek() + " ");
Queue.remove();
}
}
public static void reverseQueue(Queue<Integer> q)
{
if (q.size() == 0)
return;
int fr = q.peek();
q.remove();
reverseQueue(q);
q.add(fr);
}
public static void main(String[] args)
{
Queue<Integer> Queue = new LinkedList<>();
Queue.add(10);
Queue.add(20);
Queue.add(30);
Queue.add(40);
Queue.add(50);
Queue.add(60);
Queue.add(70);
Queue.add(80);
Queue.add(90);
Queue.add(100);
reverseQueue(Queue);
Print(Queue);
}
}
|
Python3
def Print(Queue):
while (len(Queue) > 0):
print(Queue[0],end = " ")
Queue.pop(0)
def reverseQueue(q):
if (len(q) == 0):
return
fr = q[0]
q.pop(0)
reverseQueue(q)
q.append(fr)
Queue = []
Queue.append(10)
Queue.append(20)
Queue.append(30)
Queue.append(40)
Queue.append(50)
Queue.append(60)
Queue.append(70)
Queue.append(80)
Queue.append(90)
Queue.append(100)
reverseQueue(Queue)
Print(Queue)
|
C#
using System;
using System.Collections;
public class GFG {
public static void Print(Queue Queue)
{
while (Queue.Count > 0) {
Console.Write(Queue.Peek());
Console.Write(" ");
Queue.Dequeue();
}
}
public static void reverseQueue(Queue q)
{
if (q.Count == 0)
return;
int fr = (int)q.Peek();
q.Dequeue();
reverseQueue(q);
q.Enqueue(fr);
}
static public void Main()
{
Queue Queue = new Queue();
Queue.Enqueue(10);
Queue.Enqueue(20);
Queue.Enqueue(30);
Queue.Enqueue(40);
Queue.Enqueue(50);
Queue.Enqueue(60);
Queue.Enqueue(70);
Queue.Enqueue(80);
Queue.Enqueue(90);
Queue.Enqueue(100);
reverseQueue(Queue);
Print(Queue);
}
}
|
Javascript
function Print(Queue) {
while (Queue.length != 0) {
console.log(Queue[0]);
Queue.shift();
}
}
function reverseQueue(q) {
if (q.length == 0)
return;
let fr = q[0];
q.shift();
reverseQueue(q);
q.push(fr);
}
let Queue = [];
Queue.push(10);
Queue.push(20);
Queue.push(30);
Queue.push(40);
Queue.push(50);
Queue.push(60);
Queue.push(70);
Queue.push(80);
Queue.push(90);
Queue.push(100);
reverseQueue(Queue);
Print(Queue);
|
Output100 90 80 70 60 50 40 30 20 10
Time Complexity: O(N).
Auxiliary Space: O(N). The recursion stack contains all elements of queue at a moment.
This article is contributed by Raghav Sharma and improved by Nakshatra Chhillar . If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.