Append the elements of queue in mirror-inverse order

Given a queue Q containing N strings, the task is to restructure the queue to double its size such that the second half represents the mirror image of the first half.

Examples:

Input: Q = {“Hello”, “World”}
Output: {“Hello”, “World”, “World”, “Hello”}
Explanation:
The second half of the output queue is the mirror image of the first half. That is:
“Hello”, “World” | “World”, “Hello”

Input: Q = {“Hi”, “Geeks”}
Output: {“Hi”, “Geeks”, “Geeks”, “Hi”}

Approach: On observing carefully, we can say that the second half of the output queue is the reverse of the first half. That is:



  1. Store the size of queue in a variable named size.
  2. Push the Queue elements into a stack without actually losing the elements. This can be achieved by using emplace().
  3. Repeat the process until size become 0.
  4. Push the elements of the stack back to the queue.

Below is the implementation of the above approach.

C++

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// C++ program to arrange the
// elements of the queue
// to the end such that
// the halves are mirror
// order of each other
#include <bits/stdc++.h>
using namespace std;
  
// Function to display
// the elements of
// the queue
void showq(queue<string> q)
{
    // Iterating through the queue
    // and printing the elements
    while (!q.empty()) {
        cout << q.front() << " ";
        q.pop();
    }
}
  
// Fuction to produce mirror elements
void mirrorQueue(queue<string>& q)
{
    int size = q.size();
  
    // Defining a stack
    stack<string> st;
  
    // Pushing the elements
    // of a queue
    // in a stack without
    // losing them
    // from the queue
    while (size--) {
        string x = q.front();
        // Push the element
        // to the end of the
        // queue
        q.emplace(x);
        // Push the element
        // into the stack
        st.push(x);
        // Remove the element
        q.pop();
    }
  
    // Appending the elements
    // from the stack
    // to the queue
    while (!st.empty()) {
        string el = st.top();
        q.push(el);
        st.pop();
    }
}
  
// Driver Code
int main()
{
    queue<string> q;
    q.push("Hello");
    q.push("World");
    mirrorQueue(q);
    showq(q);
    return 0;
}

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Output:

Hello World World Hello

Time Complexity: O(N), where N is the size of the queue.

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