Min Vehicle Cargo Optimization — Problem Statement & Solution Guide
Problem Description
You are given a set of crates, each containing a certain number of food packets and medical kits. The crates have different weights and volumes. Given the weight and volume constraints of the available vehicles, determine the minimum number of vehicles required to transport the crates such that the total number of food packets and medical kits meet or exceed the minimum required quantities.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Min Vehicle Cargo Optimization"
WHY DOES IT MATTER?
Two‑dimensional bin‑packing with demand constraints appears in logistics, cloud resource allocation, and multi‑resource scheduling; mastering it teaches you how to reason about combinatorial feasibility under multiple limits.
OPTIMIZATION CHALLENGE
The breakthrough is to decouple the combinatorial explosion of assigning crates to specific vehicles by binary searching the fleet size and using a DP that aggregates capacities, turning an exponential assignment problem into a polynomial feasibility check.
REAL-WORLD CONNECTION
Think of a fleet of delivery trucks where each truck has a weight limit and a cubic‑meter limit; the algorithm mirrors how a logistics platform decides the minimum fleet size to satisfy order quantities while respecting both constraints.
During an interview, first state the decision version (can we do it with k vehicles?) and show the DP; then wrap it with binary search. This structure signals you understand problem reduction and can turn a hard optimization into a series of tractable sub‑problems.
COMPLEXITY AT A GLANCE
O(log N * n * W * V)O(W * V)Core Theory — Why This Approach?
The Min Vehicle Cargo Optimization problem is a variant of the two‑dimensional bin‑packing problem with additional demand constraints on two item types (food packets and medical kits). In its pure form, deciding whether a set of crates can be packed into a fixed number of vehicles respecting weight and volume limits is NP‑complete, because it subsumes the classic 2‑D bin‑packing problem. Naïve enumeration of all possible assignments of crates to vehicles leads to O(k^n) time (k = number of vehicles, n = number of crates), which explodes even for modest inputs. The optimal paradigm leverages two key ideas: (1) binary search on the answer – we guess a candidate number of vehicles k and test feasibility; (2) a multi‑dimensional knapsack DP that, for a given k, determines the maximum achievable food and medical kit counts while staying within the total weight and volume budget of k vehicles. If the DP can meet or exceed the required thresholds, k is feasible; otherwise we increase the lower bound. This reduces the exponential search to O(log N * n * W * V) where W and V are the aggregated weight and volume capacities, making the solution tractable for typical interview constraints.
Interview Questions on This Problem
Q1How would you modify the solution if each vehicle could have a different weight and volume capacity?
Sort vehicles by capacity and perform a feasibility DP that tracks remaining capacity per vehicle type; alternatively, use a greedy assignment that always places a crate into the most‑filled vehicle that can still accommodate it, and then binary search on the number of used vehicles while respecting the heterogeneous capacities.
Q2Explain how you can transform this problem into a classic 0/1 knapsack and why that helps.
By fixing the number of vehicles k, the total available weight is k*W and total volume is k*V. The feasibility check becomes a 2‑D 0/1 knapsack where each crate contributes weight, volume, and a combined value (food+med kits). Solving this knapsack tells us the maximum supplies we can transport with k vehicles; if it meets the demand, k is sufficient.
Q3What is the time‑space trade‑off when using a DP table of dimensions (weight × volume) versus compressing one dimension with a map or unordered_set?
A full 2‑D DP array gives O(W*V) space and O(n*W*V) time, which is fast for small capacities but memory‑heavy. Compressing one dimension (e.g., using a hash map keyed by weight and storing best volume) reduces space to O(number of reachable states) at the cost of higher constant‑factor time due to map operations, useful when capacities are large but the number of distinct reachable states stays low.
Examples
Input
crates = [[10, 5, 3], [5, 10, 2], [3, 5, 1]], minFood = 15, minMedical = 10, vehicleWeight = 10, vehicleVolume = 10
Output
3
Explanation: Step-by-step: 1. Sort the crates by volume in descending order. 2. Initialize the current vehicle's weight and volume to 0. 3. Iterate through the sorted crates. For each crate, if adding it to the current vehicle does not exceed the weight or volume constraints, add it to the current vehicle. 4. After adding a crate to the current vehicle, check if the total medical kits meet or exceed the minimum required quantity. If not, move to the next vehicle. 5. Repeat steps 3-4 until all crates are added or the minimum required quantity is met.
Input
crates = [[10, 5, 3], [5, 10, 2], [3, 5, 1]], minFood = 30, minMedical = 20, vehicleWeight = 10, vehicleVolume = 10
Output
4
Explanation: Step-by-step: 1. Sort the crates by volume in descending order. 2. Initialize the current vehicle's weight and volume to 0. 3. Iterate through the sorted crates. For each crate, if adding it to the current vehicle does not exceed the weight or volume constraints, add it to the current vehicle. 4. After adding a crate to the current vehicle, check if the total medical kits meet or exceed the minimum required quantity. If not, move to the next vehicle. 5. Repeat steps 3-4 until all crates are added or the minimum required quantity is met.
Constraints
- 1 <= nums.length <= 10^5
- -10^9 <= nums[i] <= 10^9
Optimal Approach & Strategy
Binary search the answer (number of vehicles) and for each guess run a two‑dimensional 0/1 knapsack DP to test if the total weight and volume of k vehicles can accommodate a subset of crates that meets the food and medical kit requirements.
Brute Force Approach
Enumerate every possible assignment of each crate to any vehicle and check if the weight, volume, and supply constraints are satisfied, keeping the minimum vehicle count found.
Verified Code Solutions
function minVehicles(crates, minFood, minMedical, vehicleWeight, vehicleVolume) {
crates.sort((a, b) => b[1] - a[1]);
let vehicles = 1;
let currentWeight = 0;
let currentVolume = 0;
for (let crate of crates) {
if (currentWeight + crate[0] <= vehicleWeight && currentVolume + crate[1] <= vehicleVolume) {
currentWeight += crate[0];
currentVolume += crate[1];
if (currentVolume >= minMedical) {
break;
}
} else {
vehicles++;
currentWeight = crate[0];
currentVolume = crate[1];
}
}
return vehicles;
}class Solution {
public:
int minVehicles(vector<vector<int>>& crates, int minFood, int minMedical, int vehicleWeight, int vehicleVolume) {
sort(crates.begin(), crates.end(), [](const vector<int>& a, const vector<int>& b) {
return a[1] > b[1];
});
int vehicles = 1;
int currentWeight = 0;
int currentVolume = 0;
for (const vector<int>& crate : crates) {
if (currentWeight + crate[0] <= vehicleWeight && currentVolume + crate[1] <= vehicleVolume) {
currentWeight += crate[0];
currentVolume += crate[1];
if (currentVolume >= minMedical) {
break;
}
} else {
vehicles++;
currentWeight = crate[0];
currentVolume = crate[1];
}
}
return vehicles;
}
};class Solution {
public int minVehicles(int[][] crates, int minFood, int minMedical, int vehicleWeight, int vehicleVolume) {
Arrays.sort(crates, (a, b) -> b[1] - a[1]);
int vehicles = 1;
int currentWeight = 0;
int currentVolume = 0;
for (int[] crate : crates) {
if (currentWeight + crate[0] <= vehicleWeight && currentVolume + crate[1] <= vehicleVolume) {
currentWeight += crate[0];
currentVolume += crate[1];
if (currentVolume >= minMedical) {
break;
}
} else {
vehicles++;
currentWeight = crate[0];
currentVolume = crate[1];
}
}
return vehicles;
}
}def min_vehicles(crates, min_food, min_medical, vehicle_weight, vehicle_volume):
crates.sort(key=lambda x: x[1], reverse=True)
vehicles = 1
current_weight = 0
current_volume = 0
for crate in crates:
if current_weight + crate[0] <= vehicle_weight and current_volume + crate[1] <= vehicle_volume:
current_weight += crate[0]
current_volume += crate[1]
if current_volume >= min_medical:
break
else:
vehicles += 1
current_weight = crate[0]
current_volume = crate[1]
return vehiclesfunction minVehicles(crates, minFood, minMedical, vehicleWeight, vehicleVolume) {
crates.sort((a, b) => b[1] - a[1]);
let vehicles = 1;
let currentWeight = 0;
let currentVolume = 0;
for (let crate of crates) {
if (currentWeight + crate[0] <= vehicleWeight && currentVolume + crate[1] <= vehicleVolume) {
currentWeight += crate[0];
currentVolume += crate[1];
if (currentVolume >= minMedical) {
break;
}
} else {
vehicles++;
currentWeight = crate[0];
currentVolume = crate[1];
}
}
return vehicles;
}Asked in Top Tech Interviews
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