Implement Stack Using Queue in DSA C++ - Time & Space Complexity

Time Complexity: Implement Stack Using Queue

O(n)

Understanding Time Complexity

We want to understand how fast the stack operations run when we build a stack using a queue.

Specifically, how does the time to push or pop items grow as we add more elements?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

#include <queue>
using namespace std;

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();
    }
  }
  void pop() {
    q.pop();
  }
  int top() {
    return q.front();
  }
  bool empty() {
    return q.empty();
  }
};
    

This code uses one queue to simulate a stack. The push operation rearranges the queue so the newest element is always at the front.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: The for-loop inside the push method that moves elements in the queue.
• How many times: It runs (n - 1) times where n is the current size of the queue.

How Execution Grows With Input

Each push moves all existing elements behind the new one, so the work grows as the stack grows.

Input Size (n)	Approx. Operations in push()
10	9 moves
100	99 moves
1000	999 moves

Pattern observation: The number of moves grows linearly with the number of elements already in the stack.

Final Time Complexity

Time Complexity: O(n)

This means pushing an element takes longer as the stack grows, roughly proportional to the stack size.

Common Mistake

[X] Wrong: "Push operation is always O(1) because queue push is O(1)."

[OK] Correct: Here, push rearranges the queue by moving all existing elements, so it takes O(n) time, not constant time.

Interview Connect

Understanding how to build one data structure using another and analyzing its cost shows your problem-solving and reasoning skills.

Self-Check

"What if we used two queues instead of one? How would the time complexity of push and pop change?"

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