Cargo Stack Operations — Problem Statement & Solution Guide

StackMediumMixed
TimeO(N)
|
SpaceO(M)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Stack and solve the Cargo Stack Operations problem optimally.

TopicStack
PatternMixed
TimeO(N)
SpaceO(M)

Problem Description

You are given an initial stack of integers and a sequence of commands that modify the stack. The stack follows the standard LIFO rule: the most recently pushed element is removed first. The input provides the initial content of the stack (from bottom to top) followed by a list of commands. Each command is either "import x" – push the integer x onto the top of the stack, or "export" – pop the top element. If an "export" command is executed when the stack is empty, it has no effect. After processing all commands, output the remaining stack elements from bottom to top separated by a single space. If the stack is empty, output the word "EMPTY".

DSA Pattern Breakdown

DSA Pattern Breakdown

"Cargo Stack Operations"

medium

WHY DOES IT MATTER?

Stacks model many real‑world LIFO scenarios—function call frames, undo buffers, and container loading. Mastering constant‑time push/pop ensures you can meet strict latency SLAs in systems like order‑matching engines or packet processing pipelines.

OPTIMIZATION CHALLENGE

The key insight is to avoid any linear‑time reshuffling of elements. By treating the stack as a simple pointer‑based structure, each command becomes a single pointer move, turning a potentially quadratic process into a linear one.

REAL-WORLD CONNECTION

Think of a warehouse loading dock where the last container placed on a pallet is the first one taken off for delivery. Managing that pallet efficiently mirrors stack operations, and any delay in moving containers translates directly to increased turnaround time.

In an interview, write the push/pop logic with a pre‑allocated int[] and a top index; it signals to the interviewer that you understand memory layout, avoid hidden overhead, and can reason about worst‑case performance.

COMPLEXITY AT A GLANCE

⏱ Time:O(N)
💾 Space:O(M)

Core Theory — Why This Approach?

A stack is a linear data structure that follows the Last‑In‑First‑Out (LIFO) principle, meaning the most recently added element is the first one removed. In the "Cargo Stack Operations" problem the only mutable actions are push ("import x") and pop ("export"), both of which can be performed in constant time when the stack is represented with a dynamic array or linked list. Naïve solutions that repeatedly shift elements, rebuild the stack from scratch after each command, or use recursion for every pop quickly degrade to O(n²) time because each operation may touch the entire collection, which is unacceptable for large input sizes (up to 10⁶ commands).

The optimal paradigm leverages the intrinsic O(1) amortized cost of push and pop on a properly implemented stack. By maintaining a pointer (or index) to the current top, each command translates to a single array assignment or decrement, avoiding any traversal or copying. This yields linear overall time proportional to the number of commands, while memory usage stays proportional to the maximum stack height, which is also linear. The approach scales gracefully even when the command stream is streamed from disk or network, because each operation is independent and stateless beyond the top pointer.

Beyond raw performance, the stack abstraction simplifies reasoning about program state, enabling easy detection of underflow (export on an empty stack) and overflow (import beyond allocated capacity, which can be handled by dynamic resizing). These guarantees are crucial in real‑time systems where deterministic latency is required, and they form the basis for many higher‑level algorithms such as expression evaluation, backtracking, and undo mechanisms.

Interview Questions on This Problem

Q1How would you detect an underflow condition while processing the command list, and what would you output?

Maintain a size counter; before executing an "export" command, check if the counter is zero. If it is, the stack is empty—return an error code, print "EMPTY", or ignore the command based on the specification.

Q2If the input contains up to 10⁶ commands, why is using Java's Stack class (which extends Vector) potentially sub‑optimal?

Vector synchronizes every method, adding unnecessary locking overhead. A plain ArrayDeque or a custom int[] with a top index provides the same O(1) operations without the synchronization cost, yielding better throughput for large inputs.

Q3Can you extend the solution to support a "max" operation that returns the current maximum element in O(1) time? How?

Maintain a secondary stack that mirrors the main stack but stores the maximum seen so far. On each "import x", push max(x, currentMax) onto the auxiliary stack; on "export", pop from both stacks. The top of the auxiliary stack always holds the current maximum, enabling O(1) queries.

Examples

Example 1

Input

3
5 2 8
5
export
import 10
export
export
export

Output

EMPTY

Explanation: Initial stack bottom→top: [5,2,8]. export removes 8 → [5,2]. import 10 pushes 10 → [5,2,10]. export removes 10 → [5,2]. export removes 2 → [5]. export removes 5 → [] so the result is EMPTY.

Example 2

Input

0

4
import -3
import 7
export
import 2

Output

-3 2

Explanation: The stack starts empty. import -3 → [-3]. import 7 → [-3,7]. export removes 7 → [-3]. import 2 → [-3,2]. The final stack printed bottom to top is "-3 2".

Example 3

Input

2
100 200
3
export
export
export

Output

EMPTY

Explanation: Start with [100,200]. First export removes 200 → [100]. Second export removes 100 → []. Third export on an empty stack does nothing. The stack ends empty, so output is EMPTY.

Constraints

  • 1 <= initialStackSize <= 10^5
  • 0 <= numberOfCommands <= 10^5
  • -10^9 <= stack elements, x in import command <= 10^9
  • Total number of operations (initial elements + commands) does not exceed 2*10^5

Optimal Approach & Strategy

Use a dynamic array (or built‑in stack) with a top pointer; push increments the pointer and stores the value, pop decrements the pointer—both O(1).

Brute Force Approach

Repeatedly shift the entire array on every "import" or "export", which makes each operation O(n) and leads to O(n²) total time for n commands.

Code Solutions

JavaScript Solution
Time: O(N)
const fs = require('fs');
const input = fs.readFileSync(0, 'utf8').trim().split(/\s+/);
let idx = 0;
const n = Number(input[idx++]);
const init = [];
for(let i=0;i<n;i++) init.push(Number(input[idx++]));
class CargoStack {
    constructor(arr) { this.stack = [...arr]; }
    importCommand(x) { this.stack.push(x); }
    exportCommand() { return this.stack.pop(); }
    isEmpty() { return this.stack.length===0; }
}
const cs = new CargoStack(init);
const q = Number(input[idx++]);
let out = [];
for(let i=0;i<q;i++) {
    const cmd = input[idx++];
    if(cmd==='import') {
        const x = Number(input[idx++]);
        cs.importCommand(x);
    } else { // export
        if(cs.isEmpty()) out.push('EMPTY');
        else out.push(String(cs.exportCommand()));
    }
}
console.log(out.join('\n'));

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