Minimize Storage Units — Problem Statement & Solution Guide
Problem Description
A logistics center manages a fleet of modular storage containers. Each container is pre-filled with a specific, fixed quantity of three essential resources: oxygen, water, and food. The quantities vary per container and are defined by a 3-element array [oxygen, water, food].
Your task is to determine the minimum number of distinct containers required to satisfy a daily operational demand. The demand is specified by three integers: oxygenRequired, waterRequired, and foodRequired. A set of containers is considered valid if the sum of oxygen in the selected containers is at least oxygenRequired, the sum of water is at least waterRequired, and the sum of food is at least foodRequired.
Return the minimum count of containers needed to meet all three demands simultaneously. If no combination of the available containers can satisfy the requirements, return -1.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Minimize Storage Units"
WHY DOES IT MATTER?
This pattern is essential for resource allocation problems in distributed systems, cloud computing, and logistics. It teaches candidates how to handle multi-dimensional constraints in optimization problems, moving beyond simple 1D knapsack to real-world scenarios where resources are heterogeneous.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the problem is an unbounded knapsack in 3 dimensions. The optimization lies in the DP state definition: instead of tracking which containers are used, track the remaining demand. This reduces the state space from combinations of containers to the range of possible demand values.
REAL-WORLD CONNECTION
Analogous to cloud auto-scaling groups where instances have different CPU/RAM/Storage profiles, and you need to spin up the minimum number of instances to handle a specific load profile. It is also similar to cargo loading in shipping containers where weight, volume, and fragility are all constraints.
In interviews, clarify the bounds of the demand values. If the demands are large (e.g., >1000), a 3D DP array might be too large. In such cases, discuss memoization with a hash map or switch to a BFS approach on the state space, or mention that the problem becomes NP-hard and requires approximation if dimensions or values are unbounded.
COMPLEXITY AT A GLANCE
O(D_o * D_w * D_f * N)O(D_o * D_w * D_f)Core Theory — Why This Approach?
The problem of minimizing the number of distinct containers to meet specific resource demands is a variant of the Unbounded Knapsack Problem or, more specifically, a Multi-dimensional Integer Linear Programming (ILP) problem. In a naive setting, one might attempt to iterate through all possible combinations of containers, but this leads to exponential time complexity $O(N^D)$ where $N$ is the number of container types and $D$ is the dimensionality (3 in this case). For large inputs, this brute-force approach is computationally infeasible. The optimal paradigm relies on dynamic programming (DP) or, if the demands are small, a BFS/DFS on the state space of remaining demands. However, given the 'medium' difficulty and the specific structure of 'distinct containers' (implying we can use multiple units of the same container type), the problem often reduces to finding the minimum number of items to reach a target sum in a multi-dimensional space. If the container types are limited and demands are moderate, a DP table dp[oxygen][water][food] storing the minimum containers needed to satisfy those exact remaining demands is the standard approach. The recurrence relation is dp[o][w][f] = 1 + min(dp[o - c.o][w - c.w][f - c.f]) for all container types c that can fit within the current demand.
Interview Questions on This Problem
Q1At a fintech platform like Stripe, how would you model the allocation of server resources (CPU, Memory, Storage) to minimize the number of physical servers required for a batch of microservices, given that each server has fixed capacity profiles?
This is a multi-dimensional bin packing or knapsack variant. You would define the state as the remaining capacity of the current server or the total unmet demand. Using DP, you iterate through service requirements and update the minimum server count. The key is to handle the multi-dimensional constraints (CPU, Mem, Storage) simultaneously in the DP state transition, ensuring that no dimension exceeds the server's limit.
Q2In a high-growth logistics startup, if you have a fleet of drones with different payload capacities for three types of packages (light, medium, heavy), how do you minimize the number of drone flights to deliver a specific order?
Treat each drone type as an item in an unbounded knapsack where the 'value' is 1 (one flight) and the 'weight' is the vector of package counts it can carry. The goal is to reach the target demand vector with the minimum sum of 'values'. A DP approach where the state is the remaining package counts for each type allows you to compute the minimum flights by checking all possible drone types that can be dispatched next.
Q3For a product company like Amazon, how would you optimize warehouse picking routes if each picker can carry a fixed mix of item sizes (small, medium, large) and you need to fulfill an order with specific counts of each size?
This maps to minimizing the number of pickers (or trips) to satisfy the order. Each picker type has a capacity vector. The problem is to find the minimum number of pickers such that the sum of their capacity vectors is >= the order demand. This is a multi-dimensional covering problem. DP is suitable if the demand counts are bounded; otherwise, heuristic or approximation algorithms might be needed, but for interview purposes, the DP formulation min_trips[small][medium][large] is the expected solution.
Examples
Input
containers = [[5, 10, 15], [10, 5, 5], [2, 2, 2]], oxygenRequired = 12, waterRequired = 12, foodRequired = 18
Output
2
Explanation: We evaluate combinations of size 1 and size 2. No single container meets all requirements (e.g., container 0 has 5 oxygen < 12). Checking pairs: Container 0 + Container 1 yields Oxygen: 5+10=15 (>=12), Water: 10+5=15 (>=12), Food: 15+5=20 (>=18). This combination is valid. Since no single container works, the minimum is 2.
Input
containers = [[1, 1, 1], [1, 1, 1], [1, 1, 1]], oxygenRequired = 5, waterRequired = 5, foodRequired = 5
Output
-1
Explanation: The total available resources across all 3 containers are Oxygen: 3, Water: 3, Food: 3. Since the total oxygen (3) is less than the required oxygen (5), it is impossible to meet the demand regardless of the selection. Thus, return -1.
Input
containers = [[100, 0, 0], [0, 100, 0], [0, 0, 100]], oxygenRequired = 50, waterRequired = 50, foodRequired = 50
Output
3
Explanation: Container 0 provides only oxygen. Container 1 provides only water. Container 2 provides only food. To meet the oxygen requirement of 50, we must select Container 0. To meet the water requirement of 50, we must select Container 1. To meet the food requirement of 50, we must select Container 2. No subset of size 1 or 2 can cover all three distinct resource types. Therefore, all 3 containers are required.
Input
containers = [[10, 10, 10]], oxygenRequired = 5, waterRequired = 5, foodRequired = 5
Output
1
Explanation: The single container provides 10 units of each resource. Since 10 >= 5 for oxygen, water, and food, the single container satisfies all requirements. The minimum number is 1.
Constraints
- 1 <= containers.length <= 100
- 0 <= containers[i][j] <= 10^4
- 1 <= oxygenRequired, waterRequired, foodRequired <= 10^6
Optimal Approach & Strategy
Use dynamic programming with a 3D array where each cell dp[o][w][f] stores the minimum containers needed to satisfy the remaining demand o, w, and f. Iterate through all container types for each state to find the optimal transition, reducing the time complexity to polynomial in the size of the demand values.
Brute Force Approach
Recursively try every possible combination of containers, adding one container at a time, and track the minimum number of containers used to meet or exceed all three resource demands. This approach explores the entire search space of container combinations, leading to exponential time complexity.
Verified Code Solutions
function solution(storageUnits, oxygenRequired, waterRequired, foodRequired) {
let minUnits = Infinity;
for (let i = 0; i < (1 << storageUnits.length); i++) {
let oxygen = 0, water = 0, food = 0;
for (let j = 0; j < storageUnits.length; j++) {
if ((i & (1 << j)) !== 0) {
oxygen += storageUnits[j][0];
water += storageUnits[j][1];
food += storageUnits[j][2];
}
}
if (oxygen >= oxygenRequired && water >= waterRequired && food >= foodRequired) {
let units = 0;
for (let j = 0; j < storageUnits.length; j++) {
if ((i & (1 << j)) !== 0) units++;
}
minUnits = Math.min(minUnits, units);
}
}
return minUnits === Infinity ? -1 : minUnits;
}class Solution {
public:
int solution(vector<vector<int>>& storageUnits, int oxygenRequired, int waterRequired, int foodRequired) {
int minUnits = INT_MAX;
for (int i = 0; i < (1 << storageUnits.size()); i++) {
int oxygen = 0, water = 0, food = 0;
for (int j = 0; j < storageUnits.size(); j++) {
if ((i & (1 << j)) != 0) {
oxygen += storageUnits[j][0];
water += storageUnits[j][1];
food += storageUnits[j][2];
}
}
if (oxygen >= oxygenRequired && water >= waterRequired && food >= foodRequired) {
int units = 0;
for (int j = 0; j < storageUnits.size(); j++) {
if ((i & (1 << j)) != 0) units++;
}
minUnits = min(minUnits, units);
}
}
return minUnits == INT_MAX ? -1 : minUnits;
}
};class Solution {
public int solution(int[][] storageUnits, int oxygenRequired, int waterRequired, int foodRequired) {
int minUnits = Integer.MAX_VALUE;
for (int i = 0; i < (1 << storageUnits.length); i++) {
int oxygen = 0, water = 0, food = 0;
for (int j = 0; j < storageUnits.length; j++) {
if ((i & (1 << j)) != 0) {
oxygen += storageUnits[j][0];
water += storageUnits[j][1];
food += storageUnits[j][2];
}
}
if (oxygen >= oxygenRequired && water >= waterRequired && food >= foodRequired) {
int units = 0;
for (int j = 0; j < storageUnits.length; j++) {
if ((i & (1 << j)) != 0) units++;
}
minUnits = Math.min(minUnits, units);
}
}
return minUnits == Integer.MAX_VALUE ? -1 : minUnits;
}
}def solution(storageUnits, oxygenRequired, waterRequired, foodRequired):
minUnits = float('inf')
for i in range(1 << len(storageUnits)):
oxygen, water, food = 0, 0, 0
for j in range(len(storageUnits)):
if (i & (1 << j)) != 0:
oxygen += storageUnits[j][0]
water += storageUnits[j][1]
food += storageUnits[j][2]
if oxygen >= oxygenRequired and water >= waterRequired and food >= foodRequired:
units = 0
for j in range(len(storageUnits)):
if (i & (1 << j)) != 0: units += 1
minUnits = min(minUnits, units)
return -1 if minUnits == float('inf') else minUnitsfunction solution(storageUnits, oxygenRequired, waterRequired, foodRequired) {
let minUnits = Infinity;
for (let i = 0; i < (1 << storageUnits.length); i++) {
let oxygen = 0, water = 0, food = 0;
for (let j = 0; j < storageUnits.length; j++) {
if ((i & (1 << j)) !== 0) {
oxygen += storageUnits[j][0];
water += storageUnits[j][1];
food += storageUnits[j][2];
}
}
if (oxygen >= oxygenRequired && water >= waterRequired && food >= foodRequired) {
let units = 0;
for (let j = 0; j < storageUnits.length; j++) {
if ((i & (1 << j)) !== 0) units++;
}
minUnits = Math.min(minUnits, units);
}
}
return minUnits === Infinity ? -1 : minUnits;
}Asked in Top Tech Interviews
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