Given an integer k and a queue of integers, The task is to reverse the order of the first k elements of the queue, leaving the other elements in the same relative order.
Only following standard operations are allowed on queue.
- enqueue(x) : Add an item x to rear of queue
- dequeue() : Remove an item from front of queue
- size() : Returns number of elements in queue.
- front() : Finds front item.
Approach:
We can use recursive call stack and we can add remaining items of front without using additional queue.
Below are the steps:
1. Reverse first k elements.
2. Remove from front and add to back (N – K) elements.
Below is the implementation of above approach:
C++
#include <bits/stdc++.h>using namespace std;void solve(queue<int>& q, int k);// Function to reverse first k elements of a queuequeue<int> reverseFirstK(queue<int> q, int k) { solve(q, k); int s = q.size() - k; while (s-- > 0) { int x = q.front(); q.pop(); q.push(x); } return q;}void solve(queue<int>& q, int k) { if (k == 0) return; int e = q.front(); q.pop(); solve(q, k - 1); q.push(e);}// Driver codeint 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); int k = 5; queue = reverseFirstK(queue, k); // Printing queue while (!queue.empty()) { cout << queue.front() << " "; queue.pop(); } return 0;} |
Java
import java.io.*;import java.util.*;public class GFG { // Function to reverse first k elements of a queue. static Queue<Integer> reverseFirstK(Queue<Integer> q, int k) { solve(q, k); int s = q.size() - k; while( s-- > 0){ int x = q.poll(); q.add(x); } return q; } static void solve(Queue<Integer> q, int k){ if(k == 0) return; int e = q.poll(); solve(q, k - 1); q.add(e); } // driver code public static void main (String[] args) { Queue<Integer> 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); int k = 5; queue = reverseFirstK(queue, k); // printing queue while (!queue.isEmpty()) { System.out.print(queue.poll() + " "); } }} |
Python3
from collections import dequedef reverse_first_k(q, k): solve(q, k) s = len(q) - k for _ in range(s): x = q.popleft() q.append(x) return qdef solve(q, k): if k == 0: return e = q.popleft() solve(q, k - 1) q.append(e)# Driver codequeue = deque([10, 20, 30, 40, 50, 60, 70, 80, 90, 100])k = 5queue = reverse_first_k(queue, k)# Printing queuewhile queue: print(queue.popleft(), end=' ') |
C#
using System;using System.Collections.Generic;class GFG { public static LinkedList<int> queue; public static void solve(int k) { if (k == 0){ return; } int e = queue.First.Value; queue.RemoveFirst(); solve(k - 1); queue.AddLast(e); } // Function to reverse first k elements of a queue public static void reverseFirstK(int k) { solve(k); int s = queue.Count - k; while (s > 0) { int x = queue.First.Value; queue.RemoveFirst(); queue.AddLast(x); s = s - 1; } } // Driver code 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); int k = 5; reverseFirstK(k); // Printing queue while (queue.Count > 0) { Console.Write(queue.First.Value + " "); queue.RemoveFirst(); } }}// The code is contributed by Gautam goel. |
Javascript
// JavaScript Code to implement queue data structure in Javascriptclass Queue { constructor() { this.items = []; } // add element to the queue push(element) { return this.items.push(element); } // remove element from the queue pop() { if (this.items.length > 0) { return this.items.shift(); } } // view the first element front() { return this.items[0]; } // check if the queue is empty isEmpty() { return this.items.length == 0; } // the size of the queue size() { return this.items.length; }}// Function to reverse first k elements of a queuefunction reverseFirstK(queue, k) { solve(queue, k); let s = queue.size() - k; while (s-- > 0) { let x = queue.front(); queue.pop(); queue.push(x); } return queue;}function solve(queue, k) { if (k == 0) return; let e = queue.front(); queue.pop(); solve(queue, k - 1); queue.push(e);}// Driver codelet queue = new 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);let k = 5;q = reverseFirstK(queue, k);// Printing queuewhile (!q.isEmpty()) { console.log(q.front()); q.pop();}// This code is contributed by satwiksuman. |
Output
50 40 30 20 10 60 70 80 90 100
Time and Space complexity:
The time complexity of the given program can be analyzed as follows:
The function reverseFirstK calls the recursive function solve, which takes O(k) time to reverse the first k elements of the queue.
The remaining part of the function reverseFirstK takes O(n-k) time to move the remaining elements to the end of the queue.
The overall time complexity of the function reverseFirstK is O(n), where n is the size of the input queue.
Therefore, the time complexity of the entire program is O(n).
The space complexity of the program is also O(n), as the input queue is stored in memory along with some additional variables used in the program, such as the integer variable s. However, the space used by the recursive function solve is O(k), as it calls itself recursively k times, where k is the number of elements to be reversed.
Therefore, the overall space complexity of the program is O(n+k).
Approach:
The idea is to use an auxiliary stack. Store the first k elements of the queue in a stack and pop it from the queue, then push it back to the queue and perform pop operation for n-k times and again push the popped element.
Follow the below steps to implement the idea:
- Create an empty stack.
- One by one dequeue first K items from given queue and push the dequeued items to stack.
- Enqueue the contents of stack at the back of the queue
- Dequeue (size-k) elements from the front and enqueue them one by one to the same queue.
Below is the implementation of above approach:
C++
// C++ program to reverse first// k elements of a queue.#include <bits/stdc++.h>using namespace std;/* Function to reverse the first K elements of the Queue */void reverseQueueFirstKElements(int k, queue<int>& Queue){ if (Queue.empty() == true || k > Queue.size()) return; if (k <= 0) return; stack<int> Stack; /* Push the first K elements into a Stack*/ for (int i = 0; i < k; i++) { Stack.push(Queue.front()); Queue.pop(); } /* Enqueue the contents of stack at the back of the queue*/ while (!Stack.empty()) { Queue.push(Stack.top()); Stack.pop(); } /* Remove the remaining elements and enqueue them at the end of the Queue*/ for (int i = 0; i < Queue.size() - k; i++) { Queue.push(Queue.front()); Queue.pop(); }}/* Utility Function to print the Queue */void Print(queue<int>& Queue){ while (!Queue.empty()) { cout << Queue.front() << " "; Queue.pop(); }}// Driver codeint 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); int k = 5; reverseQueueFirstKElements(k, Queue); Print(Queue);} |
Java
// Java program to reverse first k elements// of a queue.import java.util.LinkedList;import java.util.Queue;import java.util.Stack;public class Reverse_k_element_queue { static Queue<Integer> queue; // Function to reverse the first // K elements of the Queue static void reverseQueueFirstKElements(int k) { if (queue.isEmpty() == true || k > queue.size()) return; if (k <= 0) return; Stack<Integer> stack = new Stack<Integer>(); // Push the first K elements into a Stack for (int i = 0; i < k; i++) { stack.push(queue.peek()); queue.remove(); } // Enqueue the contents of stack // at the back of the queue while (!stack.empty()) { queue.add(stack.peek()); stack.pop(); } // Remove the remaining elements and enqueue // them at the end of the Queue for (int i = 0; i < queue.size() - k; i++) { queue.add(queue.peek()); queue.remove(); } } // Utility Function to print the Queue static void Print() { while (!queue.isEmpty()) { System.out.print(queue.peek() + " "); queue.remove(); } } // Driver code 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); int k = 5; reverseQueueFirstKElements(k); Print(); }}// This code is contributed by Sumit Ghosh |
Python3
# Python3 program to reverse first k# elements of a queue.from queue import Queue# Function to reverse the first K# elements of the Queuedef reverseQueueFirstKElements(k, Queue): if (Queue.empty() == True or k > Queue.qsize()): return if (k <= 0): return Stack = [] # put the first K elements # into a Stack for i in range(k): Stack.append(Queue.queue[0]) Queue.get() # Enqueue the contents of stack # at the back of the queue while (len(Stack) != 0): Queue.put(Stack[-1]) Stack.pop() # Remove the remaining elements and # enqueue them at the end of the Queue for i in range(Queue.qsize() - k): Queue.put(Queue.queue[0]) Queue.get()# Utility Function to print the Queuedef Print(Queue): while (not Queue.empty()): print(Queue.queue[0], end=" ") Queue.get()# Driver codeif __name__ == '__main__': Queue = Queue() Queue.put(10) Queue.put(20) Queue.put(30) Queue.put(40) Queue.put(50) Queue.put(60) Queue.put(70) Queue.put(80) Queue.put(90) Queue.put(100) k = 5 reverseQueueFirstKElements(k, Queue) Print(Queue)# This code is contributed by PranchalK |
C#
// C# program to reverse first k elements// of a queue.using System;using System.Collections.Generic;class GFG { public static LinkedList<int> queue; // Function to reverse the first K // elements of the Queue public static void reverseQueueFirstKElements(int k) { if (queue.Count == 0 || k > queue.Count) { return; } if (k <= 0) { return; } Stack<int> stack = new Stack<int>(); // Push the first K elements into a Stack for (int i = 0; i < k; i++) { stack.Push(queue.First.Value); queue.RemoveFirst(); } // Enqueue the contents of stack at // the back of the queue while (stack.Count > 0) { queue.AddLast(stack.Peek()); stack.Pop(); } // Remove the remaining elements and // enqueue them at the end of the Queue for (int i = 0; i < queue.Count - k; i++) { queue.AddLast(queue.First.Value); queue.RemoveFirst(); } } // Utility Function to print the Queue public static void Print() { while (queue.Count > 0) { Console.Write(queue.First.Value + " "); queue.RemoveFirst(); } } // Driver code 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); int k = 5; reverseQueueFirstKElements(k); Print(); }}// This code is contributed by Shrikant13 |
Javascript
<script>// JavaScript program to reverse first// k elements of a queue./* Function to reverse the firstK elements of the Queue */function reverseQueueFirstKElements(k,Queue){ if (Queue.length == 0 || k > Queue.length) return; if (k <= 0) return; let Stack = []; /* Push the first K elements into a Stack*/ for (let i = 0; i < k; i++) { Stack.push(Queue.shift()); } /* Enqueue the contents of stack at the back of the queue*/ while (Stack.length > 0) { Queue.push(Stack.pop()); } /* Remove the remaining elements and enqueue them at the end of the Queue*/ for (let i = 0; i < Queue.length - k; i++) { Queue.push(Queue.shift()); }}/* Utility Function to print the Queue */function Print(Queue){ while (Queue.length > 0) { document.write(Queue.shift()," "); }}// Driver codelet 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);let k = 5;reverseQueueFirstKElements(k, Queue);Print(Queue);// This code is contributed by shinjanpatra</script> |
Output
50 40 30 20 10 60 70 80 90 100
Time Complexity: O(N + k), Where ‘n’ is the total number of elements in the queue and ‘k’ is the number of elements to be reversed. This is because firstly the whole queue is emptied into the stack and after that first ‘k’ elements are emptied and enqueued in the same way.
Auxiliary Space: O(k) where k is no of elements to be reversed since stack is being used to store values for the purpose of reversing.
This article is contributed by Raghav Sharma. If you like neveropen 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 neveropen main page and help other Geeks.
Using Doubly Ended Queue:
Follow the steps below to implement the above idea:
- Dequeue the first k elements of the queue and push them onto a deque using the push_front() function.
- Pop the elements from the deque one by one using the pop_front() function and enqueue them back into the queue using the push() function.
- Dequeue the remaining elements from the queue and enqueue them back into the queue using the push() and pop() functions.
Below is the implementation of the above approach:
C++
// C++ program to reverse first// k elements of a queue using dequeue.#include <deque>#include <iostream>#include <queue>using namespace std;// Function to reverse first k element of the queuevoid reverseFirstK(queue<int>& q, int k){ deque<int> d; // Dequeue the first k elements of the queue and push // them onto a deque for (int i = 0; i < k; i++) { d.push_front(q.front()); q.pop(); } // Pop the elements from the deque and enqueue them back // into the queue while (!d.empty()) { q.push(d.front()); d.pop_front(); } // Dequeue the remaining elements from the queue and // enqueue them back into the queue for (int i = 0; i < q.size() - k; i++) { q.push(q.front()); q.pop(); }}// Driver code.int main(){ queue<int> q; q.push(10); q.push(20); q.push(30); q.push(40); q.push(50); q.push(60); q.push(70); q.push(80); q.push(90); q.push(100); int k = 5; // function call. reverseFirstK(q, k); while (!q.empty()) { cout << q.front() << " "; q.pop(); } return 0;}// This code is contributed by Veerendra_Singh_Rajpoot |
Java
import java.util.*;public class ReverseFirstKQueue { // Function to reverse first k elements of the queue public static void reverseFirstK(Queue<Integer> q, int k){ Deque<Integer> d = new ArrayDeque<>(); // Dequeue the first k elements of the queue and // push them onto a deque for (int i = 0; i < k; i++) { d.push(q.poll()); } // Pop the elements from the deque and enqueue them // back into the queue while (!d.isEmpty()) { q.add(d.pop()); } // Dequeue the remaining elements from the queue and // enqueue them back into the queue for (int i = 0; i < q.size() - k; i++) { q.add(q.poll()); } } // Driver code. public static void main(String[] args){ Queue<Integer> q = new LinkedList<Integer>(); q.add(10); q.add(20); q.add(30); q.add(40); q.add(50); q.add(60); q.add(70); q.add(80); q.add(90); q.add(100); int k = 5; // function call. reverseFirstK(q, k); while (!q.isEmpty()) { System.out.print(q.peek() + " "); q.poll(); } }} |
Python3
# Python program to reverse first k elements of a queue using dequeue.from collections import dequedef reverseFirstK(q, k): d = deque() # Dequeue the first k elements of the queue and push # them onto a deque for i in range(k): d.appendleft(q.popleft()) # Pop the elements from the deque and enqueue them back # into the queue while d: q.append(d.popleft()) # Dequeue the remaining elements from the queue and # enqueue them back into the queue for i in range(len(q) - k): q.append(q.popleft())# Driver code.q = deque()q.append(10)q.append(20)q.append(30)q.append(40)q.append(50)q.append(60)q.append(70)q.append(80)q.append(90)q.append(100)k = 5# function call.reverseFirstK(q, k)print(*q) |
C#
using System;using System.Collections.Generic;public class ReverseFirstKQueue { // Function to reverse first k elements of the queue public static void reverseFirstK(Queue<int> q, int k){ Stack<int> s = new Stack<int>(); // Dequeue the first k elements of the queue and // push them onto a stack for (int i = 0; i < k; i++) { s.Push(q.Dequeue()); } // Pop the elements from the stack and enqueue them // back into the queue while (s.Count > 0) { q.Enqueue(s.Pop()); } // Dequeue the remaining elements from the queue and // enqueue them back into the queue for (int i = 0; i < q.Count - k; i++) { q.Enqueue(q.Dequeue()); } } // Driver code. public static void Main(){ Queue<int> q = new Queue<int>(); q.Enqueue(10); q.Enqueue(20); q.Enqueue(30); q.Enqueue(40); q.Enqueue(50); q.Enqueue(60); q.Enqueue(70); q.Enqueue(80); q.Enqueue(90); q.Enqueue(100); int k = 5; // function call. reverseFirstK(q, k); while (q.Count > 0) { Console.Write(q.Peek() + " "); q.Dequeue(); } }} |
Javascript
// Javascript program to reverse first// k elements of a queue using dequeue.function reverseFirstK(q, k) { const d = []; // Dequeue the first k elements of the queue and push // them onto a deque for (let i = 0; i < k; i++) { d.unshift(q.shift()); } // Pop the elements from the deque and enqueue them back // into the queue while (d.length !== 0) { q.push(d.shift()); } // Dequeue the remaining elements from the queue and // enqueue them back into the queue for (let i = 0; i < q.length - k; i++) { q.push(q.shift()); }}// Driver code.const q = [];q.push(10);q.push(20);q.push(30);q.push(40);q.push(50);q.push(60);q.push(70);q.push(80);q.push(90);q.push(100);const k = 5;// function call.reverseFirstK(q, k);console.log(q.join(' ')); |
Output
50 40 30 20 10 60 70 80 90 100
Time Complexity: O(n), The time complexity of this approach is O(n) where n is the total number of elements in the queue.
Auxiliary Space: O(k) , The space complexity of this approach is O(k) because we are using a deque to store the first k elements of the queue.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
