Queue Using Two Stacks in DSA C - Time & Space Complexity

Time Complexity: Queue Using Two Stacks

O(1) amortized

Understanding Time Complexity

We want to understand how fast operations run when we build a queue using two stacks.

Specifically, how does the time to add or remove items change as the queue grows?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

// Push element x to the back of queue
void enqueue(Stack* s1, int x) {
    push(s1, x);
}

// Remove element from front of queue
int dequeue(Stack* s1, Stack* s2) {
    if (isEmpty(s2)) {
        while (!isEmpty(s1)) {
            push(s2, pop(s1));
        }
    }
    return pop(s2);
}
    

This code uses two stacks to create a queue. Enqueue adds to the first stack. Dequeue moves all items to the second stack if empty, then pops from it.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Moving all elements from the first stack to the second stack during dequeue when the second stack is empty.
• How many times: This moving happens only when the second stack is empty, not every dequeue call.

How Execution Grows With Input

When the second stack is empty, dequeue moves all n elements once, which takes n steps. But this happens only after n enqueues.

Input Size (n)	Approx. Operations
10	About 10 moves spread over dequeues
100	About 100 moves spread over dequeues
1000	About 1000 moves spread over dequeues

Pattern observation: Although some dequeues take longer, on average each operation takes constant time because moves are spread out.

Final Time Complexity

Time Complexity: O(1) amortized

This means each enqueue or dequeue takes constant time on average, even if some dequeues take longer occasionally.

Common Mistake

[X] Wrong: "Every dequeue takes O(n) time because it moves all elements."

[OK] Correct: The moving happens only when the second stack is empty, so the cost is spread out over many dequeues, making average time constant.

Interview Connect

Understanding amortized time helps you explain why some operations seem slow but are efficient overall, a key skill in coding interviews.

Self-Check

What if we used three stacks instead of two? How would the time complexity change?

Practice

1. What is the main idea behind implementing a Queue Using Two Stacks?

easy

A. Use two stacks to store elements in sorted order

B. Use two stacks to reverse the order twice to maintain queue order

C. Use one stack for enqueue and another for dequeue operations

D. Use two stacks to perform only enqueue operations

Solution

1. Step 1: Understand queue order

   A queue follows FIFO (First In First Out) order, meaning the first element added is the first to be removed.

2. Step 2: Use two stacks to reverse order twice

   One stack reverses the order when pushing elements, and the second stack reverses it again when popping, restoring FIFO order.

3. Final Answer:

   Use two stacks to reverse the order twice to maintain queue order -> Option B

4. Quick Check:

   Two reversals keep FIFO order [OK]

Hint: Two stacks reverse order twice to simulate queue [OK]

Common Mistakes:

• Thinking both stacks store elements in sorted order
• Using stacks only for enqueue or dequeue, not both
• Confusing stack order with queue order

2. Which of the following is the correct syntax to push an element x onto a stack named stack1 in C?

easy

A. push(stack1, x);

B. stack1.push(x); // C syntax

C. stack1.push(x);

D. push(x, stack1);

Solution

1. Step 1: Recall stack push function in C

   In C, stacks are usually implemented with functions like push(stack, element).

2. Step 2: Identify correct argument order

   The stack is passed first, then the element to push, so push(stack1, x) is correct.

3. Final Answer:

   push(stack1, x); -> Option A

4. Quick Check:

   Function call with stack first, element second [OK]

Hint: In C, push(stack, element) is standard [OK]

Common Mistakes:

• Using object.method() syntax which is not valid in C
• Swapping argument order in push function
• Assuming push is a built-in method

3. Consider the following code snippet implementing a queue using two stacks s1 and s2. What will be the output after the operations?

push(s1, 10);
push(s1, 20);
pop(s2);
if (isEmpty(s2)) {
  while (!isEmpty(s1)) {
    push(s2, top(s1));
pop(s1);
  }
}
pop(s2);
printQueue(s1, s2);

medium

A. Queue elements: 10 -> null

B. Queue elements: null

C. Queue elements: 20 -> null

D. Queue elements: 10 -> 20 -> null

Solution

1. Step 1: Push 10 and 20 onto s1

   s1 = [10 (bottom), 20 (top)], s2 = []

2. Step 2: Pop from s2 (empty), then transfer all from s1 to s2

   Transfer reverses order: s2 = [20 (bottom), 10 (top)], s1 = []

3. Step 3: Pop from s2 removes 10 (top)

   s2 = [20]

4. Step 4: printQueue shows elements in queue order

   Queue order is s2 top to bottom then s1 bottom to top, so only 20 remains.

5. Final Answer:

   Queue elements: 20 -> null -> Option C

6. Quick Check:

   First popped element was 10, next is 20 [OK]

Hint: Transfer reverses stack order to simulate queue dequeue [OK]

Common Mistakes:

• Not transferring elements from s1 to s2 when s2 is empty
• Confusing top and bottom elements during transfer
• Assuming pop removes from s1 directly

4. The following code attempts to implement dequeue operation in a queue using two stacks s1 and s2. What is the bug?

int dequeue() {
  if (isEmpty(s2)) {
    while (!isEmpty(s1)) {
      push(s2, top(s1));
      pop(s1);
    }
  }
  return pop(s2);
}

medium

A. Missing check if both stacks are empty before dequeue

B. pop(s2) returns void, should use top(s2) before pop

C. Should push from s2 to s1 instead of s1 to s2

D. pop(s1) should be called after push(s2, top(s1))

Solution

1. Step 1: Check for empty queue condition

   The code does not check if both s1 and s2 are empty before dequeue, which can cause errors.

2. Step 2: Identify missing empty check

   Without this check, pop(s2) may be called on empty stack causing runtime error.

3. Final Answer:

   Missing check if both stacks are empty before dequeue -> Option A

4. Quick Check:

   Empty queue must be handled before popping [OK]

Hint: Always check if queue is empty before dequeue [OK]

Common Mistakes:

• Assuming pop returns the popped value in C without checking
• Ignoring empty queue condition
• Confusing push direction between stacks

5. You want to implement a queue using two stacks s1 and s2. You perform the following operations:

enqueue(1);
enqueue(2);
dequeue();
enqueue(3);
dequeue();
dequeue();

What will be the output sequence of the dequeued elements?

hard

A. [3, 2, 1]

B. [1, 3, 2]

C. [2, 1, 3]

D. [1, 2, 3]

Solution

1. Step 1: Enqueue 1 and 2

   s1 = [1 (bottom), 2 (top)], s2 = []

2. Step 2: Dequeue first element

   s2 empty, transfer all from s1 to s2: s2 = [2 (bottom), 1 (top)], s1 = []
   Pop from s2 removes 1

3. Step 3: Enqueue 3

   Push 3 to s1: s1 = [3], s2 = [2]

4. Step 4: Dequeue second element

   s2 not empty, pop from s2 removes 2

5. Step 5: Dequeue third element

   s2 empty, transfer s1 to s2: s2 = [3], s1 = []
   Pop from s2 removes 3

6. Final Answer:

   [1, 2, 3] -> Option D

7. Quick Check:

   FIFO order maintained by two stacks [OK]

Hint: Dequeue triggers transfer only if second stack empty [OK]

Common Mistakes:

• Not transferring elements when s2 is empty
• Assuming enqueue adds to s2 instead of s1
• Mixing order of popped elements