# Reversing the first K elements of a Queue

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 using namespace std;   void solve(queue& q, int k);   // Function to reverse first k elements of a queue queue reverseFirstK(queue 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& q, int k) {     if (k == 0) return;     int e = q.front();     q.pop();     solve(q, k - 1);     q.push(e); }   // Driver code int main() {     queue 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; }

## Python3

 from collections import deque   def reverse_first_k(q, k):     solve(q, k)     s = len(q) - k     for _ in range(s):         x = q.popleft()         q.append(x)     return q   def solve(q, k):     if k == 0:         return     e = q.popleft()     solve(q, k - 1)     q.append(e)   # Driver code queue = deque([10, 20, 30, 40, 50, 60, 70, 80, 90, 100]) k = 5 queue = reverse_first_k(queue, k)   # Printing queue while queue:     print(queue.popleft(), end=' ')

## Javascript

 // JavaScript Code to implement queue data structure in Javascript class 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 queue function 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 code   let 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 queue while (!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 using namespace std;   /* Function to reverse the first    K elements of the Queue */ void reverseQueueFirstKElements(int k, queue& Queue) {     if (Queue.empty() == true || k > Queue.size())         return;     if (k <= 0)         return;       stack 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& Queue) {     while (!Queue.empty()) {         cout << Queue.front() << " ";         Queue.pop();     } }   // Driver code int main() {     queue 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); }

## Python3

 # Python3 program to reverse first k # elements of a queue. from queue import Queue   # Function to reverse the first K # elements of the Queue     def 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 Queue     def Print(Queue):     while (not Queue.empty()):         print(Queue.queue[0], end=" ")         Queue.get()     # Driver code if __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

## Javascript



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.

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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 #include #include   using namespace std; // Function to reverse first k element of the queue void reverseFirstK(queue& q, int k) {     deque 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 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

## Python3

 # Python program to reverse first k elements of a queue using dequeue. from collections import deque   def 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 q, int k){         Stack s = new Stack();           // 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 q = new Queue();         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.

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