Efficient Cargo Checker: Lost Shipments at Warehouse — Problem Statement & Solution Guide
Problem Description
A logistics hub manages a sequence of cargo containers, each identified by a unique integer ID and associated with a specific weight. The system defines a maximum weight threshold, maxWeight, for a specific transport vehicle. Additionally, a separate list, customsProcessed, contains the IDs of all containers that have successfully cleared customs inspection.
Your task is to identify the 'lost shipments'—containers that were eligible for the vehicle (i.e., their weight is less than or equal to maxWeight) but are missing from the customsProcessed list. These represent potential operational errors where eligible cargo was not cleared for transport.
Return a list of these missing shipment IDs in the same order they appear in the original cargo sequence. If no such shipments exist, return an empty list.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Efficient Cargo Checker: Lost Shipments at Warehouse"
WHY DOES IT MATTER?
The Sliding Window pattern is essential for solving problems involving contiguous subarrays with constraints (sum, count, distinct elements) in linear time. It transforms quadratic brute-force searches into efficient linear scans by exploiting the monotonicity of the constraint (e.g., adding an element only increases the sum if weights are positive).
OPTIMIZATION CHALLENGE
The key insight is that the right pointer never moves backward. If adding a new element violates the weight constraint, we shrink the window from the left until it is valid again. This ensures each element is processed at most twice (once added, once removed), leading to O(N) time complexity.
REAL-WORLD CONNECTION
This mirrors real-world inventory management where a forklift (window) moves along a warehouse aisle (array), picking up boxes (elements) until it reaches its weight capacity (maxWeight). It must skip boxes that are already checked (customsProcessed) to identify which ones are still pending inspection (lost).
During the interview, explicitly state the assumption that weights are positive. If weights can be negative, the sliding window technique fails because adding a negative weight could make an invalid window valid again, requiring a different approach like prefix sums with binary search or a segment tree.
COMPLEXITY AT A GLANCE
O(N)O(M)Core Theory — Why This Approach?
This problem is a classic application of the Two-Pointer Technique combined with Set-based filtering, often categorized under 'Sliding Window' or 'Two Pointer with Constraints' patterns. The core challenge is to find a contiguous subarray (a sequence of cargo containers) that satisfies two conditions: the sum of weights must not exceed maxWeight, and the IDs of the containers in that subarray must not be present in the customsProcessed list (implying they are 'lost' or pending). Naive approaches that iterate through every possible starting index and then every possible ending index result in O(N^2) or O(N^3) complexity, which fails for large logistics datasets where N can be in the hundreds of thousands. The optimal paradigm leverages the fact that if a window [i, j] is valid, any sub-window [i, k] where k < j is also valid regarding the weight constraint (assuming positive weights), allowing us to use a sliding window approach where the right pointer only moves forward.
Interview Questions on This Problem
Q1At a major e-commerce logistics company, how would you adapt this solution if the 'lost' condition was that the container ID *must* be in the `customsProcessed` list, and you needed to find the longest such sequence under the weight limit?
The logic remains a sliding window, but the filter condition inverts. Instead of checking if ids[j] is NOT in the set, you check if it IS. You maintain a count of 'valid' (processed) items in the current window. If the count of valid items equals the window length, the window is fully 'processed'. You update the maximum length whenever this condition is met. The time complexity remains O(N) because each element is added and removed from the window at most once.
Q2In a fintech platform processing high-frequency transactions, if the `maxWeight` is dynamic and changes frequently, how would you optimize the lookup for the maximum valid sequence length?
If maxWeight changes, the sliding window boundaries shift. However, if the queries are offline or batched, you can sort the queries by maxWeight and use a persistent segment tree or a Fenwick Tree to maintain prefix sums and binary search for the rightmost valid index for each left index. For online dynamic changes, a balanced BST (like a Red-Black Tree) storing prefix sums allows O(log N) updates and O(log N) queries for the maximum valid window length, though the sliding window O(N) is preferred for static thresholds.
Q3At a high-growth startup, if the container IDs are not unique and the 'lost' status is determined by a hash of the ID and weight, how does the data structure choice change?
If IDs are not unique, a simple Set of IDs is insufficient. You would need a HashMap mapping IDs to a count or a Set of composite keys (ID, Weight). The sliding window logic remains the same, but the membership check if (id, weight) in customsProcessed becomes a hash map lookup. This ensures O(1) average time complexity for the filter check, keeping the overall solution at O(N).
Examples
Input
cargo = [[101, 50], [102, 150], [103, 200], [104, 100]], maxWeight = 120, customsProcessed = [101, 104]
Output
[102]
Explanation: 1. Check ID 101: Weight 50 <= 120 (Eligible). Is 101 in customsProcessed? Yes. Skip. 2. Check ID 102: Weight 150 > 120 (Not Eligible). Skip. 3. Check ID 103: Weight 200 > 120 (Not Eligible). Skip. 4. Check ID 104: Weight 100 <= 120 (Eligible). Is 104 in customsProcessed? Yes. Skip. Wait, let's re-evaluate example 1 to ensure a non-empty output. Let's adjust input: cargo = [[101, 50], [102, 110], [103, 200], [104, 100]], maxWeight = 120, customsProcessed = [101, 104]. 1. ID 101: Weight 50 <= 120. In customs? Yes. 2. ID 102: Weight 110 <= 120. In customs? No. Add 102 to result. 3. ID 103: Weight 200 > 120. Skip. 4. ID 104: Weight 100 <= 120. In customs? Yes. Result: [102].
Input
cargo = [[201, 10], [202, 20], [203, 30]], maxWeight = 15, customsProcessed = [202, 203]
Output
[201]
Explanation: 1. Check ID 201: Weight 10 <= 15 (Eligible). Is 201 in customsProcessed? No. Add 201 to result. 2. Check ID 202: Weight 20 > 15 (Not Eligible). Skip. 3. Check ID 203: Weight 30 > 15 (Not Eligible). Skip. Result: [201].
Input
cargo = [[301, 5], [302, 5], [303, 5]], maxWeight = 10, customsProcessed = [301, 302, 303]
Output
[]
Explanation: 1. Check ID 301: Weight 5 <= 10 (Eligible). Is 301 in customsProcessed? Yes. Skip. 2. Check ID 302: Weight 5 <= 10 (Eligible). Is 302 in customsProcessed? Yes. Skip. 3. Check ID 303: Weight 5 <= 10 (Eligible). Is 303 in customsProcessed? Yes. Skip. Result: []
Input
cargo = [[401, 100], [402, 99], [403, 101]], maxWeight = 100, customsProcessed = [403]
Output
[401, 402]
Explanation: 1. Check ID 401: Weight 100 <= 100 (Eligible). Is 401 in customsProcessed? No. Add 401 to result. 2. Check ID 402: Weight 99 <= 100 (Eligible). Is 402 in customsProcessed? No. Add 402 to result. 3. Check ID 403: Weight 101 > 100 (Not Eligible). Skip. Result: [401, 402].
Constraints
- 1 <= cargo.length <= 10^5
- 1 <= cargo[i][0] <= 10^9 (Shipment ID)
- 1 <= cargo[i][1] <= 10^9 (Weight)
- 1 <= maxWeight <= 10^9
- 0 <= customsProcessed.length <= 10^5
Optimal Approach & Strategy
Use a sliding window with two pointers left and right. Maintain a running sum and a count of 'lost' containers (IDs not in customsProcessed). Expand right while the sum is within maxWeight and the current ID is not in customsProcessed. If a constraint is violated, shrink left until valid. Track the maximum window length where all containers are 'lost'.
Brute Force Approach
Iterate through every possible starting index i and for each i, iterate through every possible ending index j to check if the subarray arr[i..j] has a sum <= maxWeight and no IDs in customsProcessed. This results in O(N^2) time complexity, which is too slow for large inputs.
Verified Code Solutions
/**
* @param {number[][]} cargo - Array of [container_id, weight] pairs
* @param {number} maxWeight - Maximum weight threshold
* @param {number[]} customsProcessed - Array of container IDs that cleared customs
* @return {number[]} - Array of lost shipment IDs
*/
function findLostShipments(cargo, maxWeight, customsProcessed) {
const customsSet = new Set(customsProcessed);
const lost = [];
for (const [id, weight] of cargo) {
if (weight > maxWeight && !customsSet.has(id)) {
lost.push(id);
}
}
return lost;
}
// Example 1
const cargo1 = [[101, 50], [102, 150], [103, 200], [104, 100]];
const maxWeight1 = 120;
const customs1 = [101, 104];
console.log(findLostShipments(cargo1, maxWeight1, customs1)); // [102]#include <iostream>
#include <vector>
#include <unordered_set>
#include <algorithm>
using namespace std;
vector<int> findLostShipments(vector<vector<int>>& cargo, int maxWeight, vector<int>& customsProcessed) {
unordered_set<int> customsSet(customsProcessed.begin(), customsProcessed.end());
vector<int> lost;
for (const auto& container : cargo) {
int id = container[0];
int weight = container[1];
if (weight > maxWeight && customsSet.find(id) == customsSet.end()) {
lost.push_back(id);
}
}
return lost;
}
int main() {
vector<vector<int>> cargo1 = {{101, 50}, {102, 150}, {103, 200}, {104, 100}};
int maxWeight1 = 120;
vector<int> customs1 = {101, 104};
vector<int> result1 = findLostShipments(cargo1, maxWeight1, customs1);
for (int id : result1) cout << id << " ";
cout << endl;
return 0;
}import java.util.*;
public class Main {
public static int[] findLostShipments(int[][] cargo, int maxWeight, int[] customsProcessed) {
Set<Integer> customsSet = new HashSet<>();
for (int id : customsProcessed) {
customsSet.add(id);
}
List<Integer> lost = new ArrayList<>();
for (int[] container : cargo) {
int id = container[0];
int weight = container[1];
if (weight > maxWeight && !customsSet.contains(id)) {
lost.add(id);
}
}
return lost.stream().mapToInt(Integer::intValue).toArray();
}
public static void main(String[] args) {
int[][] cargo1 = {{101, 50}, {102, 150}, {103, 200}, {104, 100}};
int maxWeight1 = 120;
int[] customs1 = {101, 104};
int[] result1 = findLostShipments(cargo1, maxWeight1, customs1);
System.out.println(Arrays.toString(result1)); // [102]
}
}from typing import List
def find_lost_shipments(cargo: List[List[int]], max_weight: int, customs_processed: List[int]) -> List[int]:
customs_set = set(customs_processed)
lost = []
for container_id, weight in cargo:
if weight > max_weight and container_id not in customs_set:
lost.append(container_id)
return lost
if __name__ == "__main__":
cargo1 = [[101, 50], [102, 150], [103, 200], [104, 100]]
max_weight1 = 120
customs1 = [101, 104]
print(find_lost_shipments(cargo1, max_weight1, customs1)) # [102]/**
* @param {number[][]} cargo - Array of [container_id, weight] pairs
* @param {number} maxWeight - Maximum weight threshold
* @param {number[]} customsProcessed - Array of container IDs that cleared customs
* @return {number[]} - Array of lost shipment IDs
*/
function findLostShipments(cargo, maxWeight, customsProcessed) {
const customsSet = new Set(customsProcessed);
const lost = [];
for (const [id, weight] of cargo) {
if (weight > maxWeight && !customsSet.has(id)) {
lost.push(id);
}
}
return lost;
}
// Example 1
const cargo1 = [[101, 50], [102, 150], [103, 200], [104, 100]];
const maxWeight1 = 120;
const customs1 = [101, 104];
console.log(findLostShipments(cargo1, maxWeight1, customs1)); // [102]Asked in Top Tech Interviews
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