Reversing the First K Items of a Queue: A Sample Problem and Solution in Easy Language

 Problem:

We have a queue of integers and an integer value k. We need to reverse the order of the first k elements in the queue while keeping the other elements in the same order.

We can only use the following standard operations on the queue:

• Enqueue(x): Add an item x to the rear of the queue
• Dequeue(): Remove an item from the front of the queue
• Size(): Returns the number of elements in the queue
• Front(): Finds the front item in the queue

Examples:

• If the queue is [10, 20, 30, 40, 50, 60, 70, 80, 90, 100] and k is 5, then the output should be [50, 40, 30, 20, 10, 60, 70, 80, 90, 100].
• If the queue is [10, 20, 30, 40, 50, 60, 70, 80, 90, 100] and k is 4, then the output should be [40, 30, 20, 10, 50, 60, 70, 80, 90, 100].

Solution:

To reverse the order of the first k elements in the queue, we can use an auxiliary stack of size k. Here are the steps:

1. Put the first k items from the front of the queue into the stack.
2. Remove the items one by one from the stack and put them back into the queue in the reverse order.
3. To arrange the first N-K elements in the queue at their required position, take them one by one from the queue, and insert them back into the queue.

Implementation:

Here's the implementation of the solution in Java:

java
public static void reverseK(Queue<Integer> q, int k) {
    if (q.size() < 0 || k < 0) {
        return;
    }
    
    Deque<Integer> stack = new ArrayDeque<Integer>();
    
    for (int i = 0; i < k; i++) {
        stack.offerLast(q.poll());
    }
    
    while (!stack.isEmpty()) {
        q.offer(stack.pollLast());
    }
        
    for (int i = 0; i < q.size() - k; i++) {
        q.offer(q.poll());
    }
}

In this code, we first check if the size of the queue or k is less than 0, then we return without doing anything. Next, we create an auxiliary stack of size k using a Deque data structure. We then add the first k items from the front of the queue to the stack using the offerLast() and poll() methods. After that, we remove items one by one from the stack and put them back into the queue using the offer() and pollLast() methods in the reverse order. Finally, we take the remaining items from the queue (N-K elements) and insert them back into the queue using the offer() and poll() methods.

I hope this explanation helps you understand the problem and its solution better!