Implement Stack Using Queue in DSA C++ - Build from Scratch

Implement Stack Using Queue

📖 Scenario: Imagine you have a stack of plates but you only have a queue (a line where people wait) to organize them. You want to use the queue to behave like a stack, where the last plate you put is the first one you take out.

🎯 Goal: You will build a stack using a single queue. You will write code to add (push) plates, remove (pop) the top plate, and see the top plate, all using queue operations.

📋 What You'll Learn

Create a queue named q using std::queue<int>

Create a function push(int x) that adds an element to the stack using the queue q

Create a function pop() that removes the top element from the stack using the queue q

Create a function top() that returns the top element of the stack using the queue q

Print the top element after pushing and popping to show the stack behavior

💡 Why This Matters

🌍 Real World

Sometimes hardware or software only provides queue operations, but you need stack behavior. This technique helps in such cases.

💼 Career

Understanding how to simulate one data structure using another is a common interview question and helps improve problem-solving skills.

1

Create the queue data structure

Create a queue named q of type int using std::queue<int>.

DSA C++

#include 
#include 

using namespace std;

int main() {
    // Create a queue named q of type int
    // Your code here

    return 0;
}

Hint

Use queue<int> q; to create the queue.

2

Add the push function to insert elements

Add a function void push(int x) that inserts x into the queue q and then rotates the queue to make the new element be at the front.

DSA C++

#include 
#include 

using namespace std;

queue q;

void push(int x) {
    // Add x to q and rotate q to put x at front
    // Your code here
}

int main() {
    queue q;

    return 0;
}

Hint

Push the element, then move all other elements behind it by popping and pushing them again.

3

Add pop and top functions

Add a function void pop() that removes the front element from q, and a function int top() that returns the front element of q.

DSA C++

#include 
#include 

using namespace std;

queue q;

void push(int x) {
    q.push(x);
    int size = q.size();
    for (int i = 0; i  q;

    return 0;
}

Hint

Use q.pop() to remove and q.front() to get the top element.

4

Test the stack operations and print the top

In main(), push 10, 20, and 30 using push. Then print the top element using top(). Next, pop one element and print the top element again.

DSA C++

#include 
#include 

using namespace std;

queue q;

void push(int x) {
    q.push(x);
    int size = q.size();
    for (int i = 0; i < size - 1; i++) {
        q.push(q.front());
        q.pop();
    }
}

void pop() {
    if (!q.empty()) {
        q.pop();
    }
}

int top() {
    if (!q.empty()) {
        return q.front();
    }
    return -1; // Return -1 if empty
}

int main() {
    // Push 10, 20, 30
    // Print top
    // Pop one element
    // Print top again
    // Your code here
    return 0;
}

Hint

Push elements in order, print top, pop once, then print top again.

Practice

1. What is the main idea behind implementing a stack using a queue?

easy

A. Use two stacks to simulate a queue

B. Use a linked list to implement both stack and queue

C. Use one queue and reorder elements to simulate stack behavior

D. Use recursion to reverse the queue elements

Solution

1. Step 1: Understand stack and queue behavior

   A stack follows Last-In-First-Out (LIFO), while a queue follows First-In-First-Out (FIFO).

2. Step 2: Simulate stack using queue

   By reordering elements in a single queue after each push, the newest element is always at the front, simulating stack pop behavior.

3. Final Answer:

   Use one queue and reorder elements to simulate stack behavior -> Option C

4. Quick Check:

   Stack via queue = reorder elements [OK]

Hint: Rearrange queue after push to mimic stack top [OK]

Common Mistakes:

• Confusing stack with queue behavior
• Using two stacks instead of queues
• Not reordering queue after push

2. Which of the following is the correct syntax to declare a queue of integers in C++ STL?

easy

A. queue q<int>;

B. queue<int> q;

C. queue<int> q()

D. queue<int> q[];

Solution

1. Step 1: Recall C++ STL queue declaration

   The correct syntax uses template syntax: queue<int> q; declares a queue of integers.

2. Step 2: Check syntax correctness

   queue<int> q; matches the correct syntax; others have syntax errors or invalid declarations.

3. Final Answer:

   queue<int> q; -> Option B

4. Quick Check:

   Correct STL queue syntax = queue<int> q; [OK]

Hint: Use template syntax queue<int> q; for integer queue [OK]

Common Mistakes:

• Misplacing template brackets
• Using parentheses instead of semicolon
• Declaring queue as array

3. What will be the output of the following code snippet implementing stack using one queue?

class Stack {
  queue<int> q;
public:
  void push(int x) {
    q.push(x);
    int n = q.size();
    for (int i = 0; i < n - 1; i++) {
      q.push(q.front());
      q.pop();
    }
  }
  int pop() {
    int val = q.front();
    q.pop();
    return val;
  }
};

int main() {
  Stack s;
  s.push(10);
  s.push(20);
  s.push(30);
  cout << s.pop() << " ";
  s.push(40);
  cout << s.pop() << " ";
  cout << s.pop() << " ";
  return 0;
}

medium

A. 30 40 20

B. 10 20 30

C. 30 20 10

D. 20 30 40

Solution

1. Step 1: Trace push operations

   Push 10: queue = [10]
   Push 20: queue = [20, 10] after rotation
   Push 30: queue = [30, 20, 10] after rotation

2. Step 2: Trace pop and push

   Pop: returns 30, queue = [20, 10]
   Push 40: queue = [40, 20, 10]
   Pop: returns 40, queue = [20, 10]
   Pop: returns 20, queue = [10]

3. Final Answer:

   30 40 20 -> Option A

4. Quick Check:

   Stack pop order = 30, 40, 20 [OK]

Hint: After push, rotate queue to front for stack order [OK]

Common Mistakes:

• Not rotating queue after push
• Confusing queue front with stack top
• Incorrect pop order

4. Identify the error in the following stack implementation using two queues:

class Stack {
  queue<int> q1, q2;
public:
  void push(int x) {
    q1.push(x);
  }
  int pop() {
    if (q1.empty()) return -1;
    while (q1.size() > 1) {
      q2.push(q1.front());
      q1.pop();
    }
    int val = q1.front();
    q1.pop();
    swap(q1, q2);
    return val;
  }
};

medium

A. Pop operation does not handle empty stack correctly

B. Push operation should push to q2 instead of q1

C. Swap should be done before popping from q1

D. Pop operation is correct; no error

Solution

1. Step 1: Analyze pop operation logic

   Pop moves all but last element from q1 to q2, then pops last element from q1, then swaps q1 and q2.

2. Step 2: Check correctness of push and pop

   Push adds to q1; pop removes last pushed element correctly by transferring elements and swapping queues.

3. Final Answer:

   Pop operation is correct; no error -> Option D

4. Quick Check:

   Two-queue stack pop logic = correct [OK]

Hint: Pop moves elements and swaps queues to simulate stack pop [OK]

Common Mistakes:

• Swapping queues too early
• Pushing to wrong queue
• Not checking empty queue before pop

5. You want to implement a stack using a single queue but optimize the pop operation to O(1) time. Which approach is best?

hard

A. Push elements by enqueueing and dequeueing all elements except the new one

B. Use a queue and a stack together for hybrid implementation

C. Use two queues to separate push and pop operations

D. Push elements normally and rotate queue on each pop

Solution

1. Step 1: Understand time complexity trade-offs

   Push elements normally and rotate queue on each pop (push normally, rotate on pop) makes push O(1), pop O(n). Use a queue and a stack together for hybrid implementation is not pure queue. Use two queues to separate push and pop operations uses two queues, not single queue.

2. Step 2: Optimize pop to O(1) with single queue

   Enqueue new element, then dequeue and enqueue all previous elements to move new element to front. This makes pop O(1) as front is always top (matches Push elements by enqueueing and dequeueing all elements except the new one).

3. Final Answer:

   Push elements by enqueueing and dequeueing all elements except the new one -> Option A

4. Quick Check:

   Rotate queue after push for O(1) pop [OK]

Hint: Rotate queue after push to keep top at front [OK]

Common Mistakes:

• Trying to optimize pop without reordering queue
• Using two queues instead of one
• Mixing stack and queue data structures