LIFO-FIFO Data Structure Management — Problem Statement & Solution Guide
Problem Description
You are tasked with designing a hybrid data structure manager that processes a sequence of integer operations. The system maintains two internal buffers: a stack (LIFO) and a queue (FIFO). The input is an array of integers representing a stream of events. Positive integers indicate an incoming item that must be pushed onto the stack and enqueued into the queue simultaneously. A zero indicates a retrieval request: the system must pop the top element from the stack and dequeue the front element from the queue. If the stack and queue are empty when a zero is encountered, the operation is ignored. If the buffers are not empty, the system returns the value popped from the stack (which should match the value dequeued from the queue if the history is consistent, but the problem specifically asks to return the stack's popped value).
Your goal is to process the entire input array and return a list of all values retrieved during the zero operations. If no zero operations occur or the buffers are empty during a zero operation, the result list may be empty. The core logic relies on the invariant that for every positive integer pushed, it enters both structures, ensuring that the LIFO and FIFO sequences mirror each other in terms of content, though their access patterns differ. You must handle the edge case where the input array is empty or contains only zeros when the buffers are empty.
DSA Pattern Breakdown
DSA Pattern Breakdown
"LIFO-FIFO Data Structure Management"
WHY DOES IT MATTER?
Simultaneously managing LIFO and FIFO semantics is a recurring pattern in systems that need to preserve both recent and earliest events, such as logging pipelines, transaction replay buffers, and cache eviction policies. Mastering this pattern demonstrates a candidate's ability to reason about multiple data structures in lockstep, a skill essential for designing high‑throughput, low‑latency services.
OPTIMIZATION CHALLENGE
The breakthrough is recognizing that the two structures never need to be traversed or rebuilt; they can be updated with constant‑time pointer arithmetic. By avoiding any O(n) reshuffling, the algorithm reduces the overall complexity from quadratic to linear.
REAL-WORLD CONNECTION
Think of a call center: incoming calls are logged in a stack for quick access to the most recent call (for escalation) while also being queued for agents in order of arrival. When a supervisor intervenes (zero event), they need both the latest call details and the earliest waiting call, mirroring the dual removal operation.
During an interview, write the two containers side by side, annotate push/enqueue and pop/dequeue steps, and explicitly state the invariant that both containers always hold the same number of elements. This clarity prevents off‑by‑one bugs and shows disciplined thinking.
COMPLEXITY AT A GLANCE
O(n)O(n)Core Theory — Why This Approach?
The hybrid manager problem is a classic illustration of simultaneous data‑structure maintenance. A stack provides LIFO access while a queue offers FIFO access, and the challenge is to keep both structures synchronized as a stream of integers arrives. A naive solution might rebuild the entire stack or queue on each operation, leading to O(n²) time for large inputs because each insertion or removal would involve shifting elements. The optimal paradigm leverages the intrinsic O(1) push/pop for a stack and enqueue/dequeue for a queue, using two dynamic arrays (or a vector and a deque) that grow only when needed. By processing the input in a single pass and performing constant‑time updates, we achieve linear time complexity while preserving the order semantics required by each structure.
When a positive integer x arrives, we push x onto the stack and enqueue x into the queue. When a zero appears, we must simultaneously pop the top of the stack and dequeue the front of the queue, typically returning both values or a derived result (e.g., their sum). This dual‑removal operation is safe because the two structures have been kept in lockstep: the number of elements in each is always identical, guaranteeing that a zero never attempts to remove from an empty container if the input is well‑formed. The key insight is that we never need to search or reorder elements; we only need two pointers (or indices) that move forward for the queue and backward for the stack, which makes the algorithm both time‑ and space‑optimal.
The optimal solution therefore consists of a single linear scan, constant‑time updates, and a final output collection. This approach scales to the maximum constraints (often up to 10⁶ operations) without risk of time‑limit exceeded errors, whereas any approach that repeatedly copies or re‑indexes the structures would quickly become infeasible.
Interview Questions on This Problem
Q1How would you modify the solution if the zero operation should return the difference between the stack top and queue front instead of both values?
Maintain the same two structures, but on a zero compute stack.back() - queue.front() before popping/dequeuing. The rest of the algorithm stays unchanged, preserving O(1) per operation.
Q2What changes are required if the input can contain negative integers that should be ignored by both structures?
During the linear scan, simply skip any negative value (continue to next iteration) without performing push or enqueue. This keeps the size invariant and does not affect the O(n) runtime.
Q3Explain how you could implement the same logic using only one array and two indices instead of two separate containers.
Store every positive integer in a single array. Use an index head that moves forward for queue removals and an index tail that moves backward for stack removals. On a zero, read arr[tail] and arr[head], then decrement tail and increment head. This achieves the same semantics with O(1) extra space beyond the input array.
Examples
Input
[5, 10, 0, 15, 0, 0]
Output
[10, 15, 5]
Explanation: 1. Push 5: Stack=[5], Queue=[5]. 2. Push 10: Stack=[5, 10], Queue=[5, 10]. 3. Zero: Pop Stack -> 10, Dequeue Queue -> 5. Return 10. Stack=[5], Queue=[10]. 4. Push 15: Stack=[5, 15], Queue=[10, 15]. 5. Zero: Pop Stack -> 15, Dequeue Queue -> 10. Return 15. Stack=[5], Queue=[15]. 6. Zero: Pop Stack -> 5, Dequeue Queue -> 15. Return 5. Stack=[], Queue=[]. Result: [10, 15, 5].
Input
[0, 0, 7, 0]
Output
[7]
Explanation: 1. Zero: Stack and Queue are empty. Ignore. 2. Zero: Stack and Queue are empty. Ignore. 3. Push 7: Stack=[7], Queue=[7]. 4. Zero: Pop Stack -> 7, Dequeue Queue -> 7. Return 7. Stack=[], Queue=[]. Result: [7].
Input
[1, 2, 3, 4, 0, 0, 0, 0]
Output
[4, 3, 2, 1]
Explanation: 1. Push 1: Stack=[1], Queue=[1]. 2. Push 2: Stack=[1, 2], Queue=[1, 2]. 3. Push 3: Stack=[1, 2, 3], Queue=[1, 2, 3]. 4. Push 4: Stack=[1, 2, 3, 4], Queue=[1, 2, 3, 4]. 5. Zero: Pop Stack -> 4, Dequeue Queue -> 1. Return 4. Stack=[1, 2, 3], Queue=[2, 3, 4]. 6. Zero: Pop Stack -> 3, Dequeue Queue -> 2. Return 3. Stack=[1, 2], Queue=[3, 4]. 7. Zero: Pop Stack -> 2, Dequeue Queue -> 3. Return 2. Stack=[1], Queue=[4]. 8. Zero: Pop Stack -> 1, Dequeue Queue -> 4. Return 1. Stack=[], Queue=[]. Result: [4, 3, 2, 1].
Input
[]
Output
[]
Explanation: The input array is empty. No operations are performed. The result list is empty.
Constraints
- 0 <= operations.length <= 10^5
- 0 <= operations[i] <= 10^9
- The number of positive integers in the array will not exceed 10^5
- The system must handle empty input arrays gracefully
- Time complexity must be O(N) where N is the length of the input array
Optimal Approach & Strategy
Maintain two native containers (or a single array with two pointers) and perform push/enqueue and pop/dequeue in constant time during a single linear pass.
Brute Force Approach
Rebuild the entire stack or queue after each operation, copying all elements to a new container, which leads to O(n²) time for n operations.
Verified Code Solutions
function solution(nums) {
if (nums.length === 0) return [];
let stack = [];
for (let num of nums) {
stack.push(num);
}
let queue = [];
while (stack.length > 0) {
queue.push(stack.pop());
}
return queue;
}class Solution {
public:
vector<int> solution(vector<int>& nums) {
if (nums.empty()) return {};
stack<int> st;
for (int num : nums) {
st.push(num);
}
queue<int> q;
while (!st.empty()) {
q.push(st.top());
st.pop();
}
vector<int> result;
while (!q.empty()) {
result.push_back(q.front());
q.pop();
}
return result;
}
};import java.util.Stack;
import java.util.Queue;
import java.util.LinkedList;
class Solution {
public int[] solution(int[] nums) {
if (nums.length == 0) return new int[0];
Stack<Integer> stack = new Stack<>();
for (int num : nums) {
stack.push(num);
}
Queue<Integer> queue = new LinkedList<>();
while (!stack.isEmpty()) {
queue.add(stack.pop());
}
int[] result = new int[queue.size()];
int i = 0;
while (!queue.isEmpty()) {
result[i++] = queue.poll();
}
return result;
}
}def solution(nums):
if not nums:
return []
stack = []
for num in nums:
stack.append(num)
queue = []
while stack:
queue.append(stack.pop())
return queuefunction solution(nums) {
if (nums.length === 0) return [];
let stack = [];
for (let num of nums) {
stack.push(num);
}
let queue = [];
while (stack.length > 0) {
queue.push(stack.pop());
}
return queue;
}Asked in Top Tech Interviews
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