Queue Using Two Stacks in DSA C - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to push an element onto the first stack.

DSA C

void enqueue(int x) {
    stack1[top1++] = [1];
}

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Atop1

Bstack2[top2]

Cx

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Using the wrong variable instead of x.

Trying to push onto stack2 instead of stack1.

We push the value x onto stack1 at the current top1 index.

2fill in blank

medium

Complete the code to check if both stacks are empty (queue is empty).

DSA C

int isEmpty() {
    return (top1 == 0 && [1] == 0);
}

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Astack2

Btop1

Cstack1

Dtop2

Attempts:

3 left

💡 Hint

Common Mistakes

Checking stack arrays instead of top indices.

Using top1 twice instead of top1 and top2.

The queue is empty if both stack1 and stack2 are empty, so top2 must be zero.

3fill in blank

hard

Fix the error in the dequeue function to correctly transfer elements from stack1 to stack2.

DSA C

int dequeue() {
    if (top2 == 0) {
        while (top1 > 0) {
            stack2[top2++] = [1];
            top1--;
        }
    }
    return stack2[--top2];
}

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Astack1[top1 - 1]

Bstack2[top2 - 1]

Cstack2[top2]

Dstack1[top1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using top1 instead of top1-1 causes out-of-bounds access.

Copying from stack2 instead of stack1.

We move the top element from stack1 (at top1-1) to stack2 and then decrease top1.

4fill in blank

hard

Fill both blanks to correctly initialize the queue and reset stack tops.

DSA C

void initQueue() {
    top1 = [1];
    top2 = [2];
}

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A0

B-1

C1

DNULL

Attempts:

3 left

💡 Hint

Common Mistakes

Setting top indices to -1 which is not used in this implementation.

Using NULL which is not valid for integer indices.

Both top1 and top2 start at 0 indicating empty stacks.

5fill in blank

hard

Fill all three blanks to correctly check if the queue is empty and return an error value if so.

DSA C

int dequeue() {
    if (top1 == 0 && top2 == 0) {
        return [1];
    }
    if (top2 == 0) {
        while (top1 > 0) {
            stack2[top2++] = stack1[[2]];
            top1--;
        }
    }
    return stack2[[3]];
}

Drag options to blanks, or click blank then click option'

A-1

Btop1 - 1

Ctop2 - 1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Returning 0 instead of -1 for empty queue.

Using top1 or top2 directly instead of top-1 for indexing.

Return -1 if empty; move from stack1[top1-1]; return stack2[top2-1].