Reversing the First K elements of a Queue

Difficulty Level Easy
Frequently asked in BlackRock JP Morgan Robinhood Sprinklr Wooker ZScaler
Queue StackViews 4811

In reversing the first K elements of a queue problem we have given a queue and a number k, reverse the first k elements of a queue using standard operations of the queue.

Examples

Input:
queue = 10 -> 15 -> 31 -> 17 -> 12 -> 19 -> 2
k = 3
Output:
queue = 31 -> 15 -> 10 -> 17 -> 12 -> 19 -> 2

Input:
queue = 12 -> 14 -> 16 -> 7 -> 9
k = 2
Output:
queue = 14 -> 12 -> 16 -> 7 -> 9

Algorithm for Reversing the First K elements of a Queue

To reverse the first k elements of the queue we can use a stack.

  1. Remove first k elements of the queue and push them into a stack.
  2. Pop all the elements of the stack and push them to the end of the queue.
  3. Pop-out (n – k) elements from the front of the queue and push them to the end of the queue, where n is the total number of elements in the queue.
  4. First, k elements of the queue are reversed, print the elements of the queue.

Explanation for Reversing the First K elements of a Queue

Consider an example,
queue = 10 -> 7 -> 4 -> 3
k = 2

Step 1

Remove first k elements of the queue and push them into a stack.
queue = 10 -> 7 -> 4 -> 3 and stack = null
Iteration 1
queue = 7 -> 4 -> 3 and stack = 10
Iteration 2
queue = 4 -> 3 and stack= 7 -> 10

Step 2

Pop all the elements of the stack and push them to the end of the queue.
queue= 4 -> 3 and stack= 7 -> 10
Iteration 1
queue= 4 -> 3 -> 7 and stack = 10
Iteration 2
queue = 4 -> 3 -> 7 -> 10 and stack = null

Step 3

Pop out (n – k) elements from the front of the queue and push them to the end of the queue
queue = 4 -> 3 -> 7 -> 10
Iteration 1
queue = 3 -> 7 -> 10 -> 4
Iteration 2
queue = 7 -> 10 -> 4 -> 3

Reversing the First K elements of a Queue

JAVA Code

import java.util.LinkedList;
import java.util.Queue;
import java.util.Stack;

public class ReversingTheFirstKElementsOfAQueue {
    private static void reverseKElements(Queue<Integer> queue, int k) {
        if (k < 0 || k >= queue.size() || queue.isEmpty()) {
            System.out.println("Invalid Input");
            return;
        }

        int n = queue.size();

        // remove first k elements of queue and push in stack
        Stack<Integer> stack = new Stack<>();
        for (int i = 0; i < k; i++) {
            int curr = queue.poll();
            stack.push(curr);
        }
        
        // Pop out elements from stack and add to the end of the queue
        while (!stack.isEmpty()) {
            int curr = stack.pop();
            queue.add(curr);
        }

        // Remove first (n - k) elements of the queue and add them to the end
        for (int i = 0; i < n - k; i++) {
            int curr = queue.poll();
            queue.add(curr);
        }

        // Print the elements of the queue
        for (Integer i : queue) {
            System.out.print(i + " ");
        }
        System.out.println();
    }

    public static void main(String[] args) {
        // Example 1
        Queue<Integer> q1 = new LinkedList<>();
        int k1 = 3;
        q1.add(10);
        q1.add(15);
        q1.add(31);
        q1.add(17);
        q1.add(12);
        q1.add(19);
        q1.add(2);
        reverseKElements(q1, k1);

        // Example 2
        Queue<Integer> q2 = new LinkedList<>();
        int k2 = 2;
        q2.add(12);
        q2.add(14);
        q2.add(16);
        q2.add(7);
        q2.add(9);
        reverseKElements(q2, k2);
    }
}
31 15 10 17 12 19 2 
14 12 16 7 9

C++ Code

#include<bits/stdc++.h> 
using namespace std;

void reverseKElements(queue<int> &queue, int k) {
    if (k < 0 || k >= queue.size() || queue.empty()) {
        cout<<"Invalid Input"<<endl;
        return;
    }
    
    int n = queue.size();
    
    // remove first k elements of queue and push in stack
    stack<int> st;
    for (int i = 0; i < k; i++) {
        int curr = queue.front();
        queue.pop();
        st.push(curr);
    }
    
    // Pop out elements from stack and add to the end of the queue
    for (int i = 0; i < k; i++) {
        int curr = st.top();
        st.pop();
        queue.push(curr);
    }
    
    // Remove first (n - k) elements of the queue and add them to the end
    for (int i = 0; i < n - k; i++) {
        int curr = queue.front();
        queue.pop();
        queue.push(curr);
    }
    
    // Print the elements of the queue
    for (int i = 0; i < n; i++) {
        int curr = queue.front();
        queue.pop();
        cout<<curr<<" ";
        queue.push(curr);
    }
    cout<<endl;
}

int main() {
    // Example 1
    queue<int> q1;
    int k1 = 3;
    q1.push(10);
    q1.push(15);
    q1.push(31);
    q1.push(17);
    q1.push(12);
    q1.push(19);
    q1.push(2);
    reverseKElements(q1, k1);

    // Example 2
    queue<int> q2;
    int k2 = 2;
    q2.push(12);
    q2.push(14);
    q2.push(16);
    q2.push(7);
    q2.push(9);
    reverseKElements(q2, k2);
}
31 15 10 17 12 19 2 
14 12 16 7 9

Complexity Analysis for Reversing the First K elements of a Queue

Time Complexity = O(n + k)
Space Complexity = O(k) 
where n is the number of elements in the queue.

References

Translate »