Optimizing Space Station Supplies — Problem Statement & Solution Guide

ArraysHardCustom

TimeO(n log n)

SpaceO(n)

Quick Answer & Algorithm Key Takeaway

Master coding challenges related to Arrays and solve the Optimizing Space Station Supplies problem optimally.

TopicArrays

PatternCustom

TimeO(n log n)

SpaceO(n)

Problem Description

Given an array of crate weights and a target sum, find the maximum total weight of supplies that can be stored in a cargo bay by selecting a subarray of crates after performing adjustments.

DSA Pattern Breakdown

"Optimizing Space Station Supplies"

hard

WHY DOES IT MATTER?

This pattern—prefix sums combined with a sorted data structure—turns a seemingly combinatorial subarray problem into a series of efficient range queries. It is a cornerstone technique for many range‑sum and maximum‑subarray variants, enabling solutions that scale to millions of elements.

OPTIMIZATION CHALLENGE

The challenge is to avoid recomputing subarray sums from scratch. By storing all prefix sums in a balanced BST, each query for the best previous prefix becomes O(log n) instead of O(n), dramatically reducing time while keeping space linear.

REAL-WORLD CONNECTION

In distributed log aggregation, you often need the largest contiguous block of logs whose size does not exceed a storage quota. Prefix sums represent cumulative log sizes, and the sorted structure allows quick identification of the optimal block, analogous to the cargo bay selection.

When implementing, always use 64‑bit integers to avoid overflow, and remember that the prefix array starts with 0 to handle subarrays that begin at index 0. Also, consider using a multiset if duplicate prefix sums are possible.

COMPLEXITY AT A GLANCE

⏱ Time:O(n log n)

💾 Space:O(n)

Core Theory — Why This Approach?

The problem reduces to finding a subarray whose sum is as large as possible without exceeding a given target. A naive approach enumerates all O(n^2) subarrays and checks their sums, which is infeasible for large n because the time grows quadratically. The optimal paradigm uses prefix sums to transform subarray sums into differences of two prefix values: sum(i..j) = prefix[j] - prefix[i-1]. By maintaining a sorted structure of all seen prefix sums, we can, for each current prefix, efficiently locate the largest earlier prefix that keeps the difference ≤ target. This reduces the problem to a series of range queries on a sorted set, solvable in O(log n) per element, yielding an overall O(n log n) algorithm. The key insight is that the subarray sum constraint translates into a simple inequality on prefix sums, enabling binary search or balanced BST usage.

Interview Questions on This Problem

Q1How would you modify the algorithm if all crate weights are guaranteed to be non‑negative?

With non‑negative weights, the subarray sums are monotonic with respect to the right endpoint. A two‑pointer sliding window can be used: expand the right pointer while the sum stays ≤ target, and when it exceeds target, shrink from the left. This achieves O(n) time because each pointer moves at most n steps.

Q2In a fintech application, you need to find the maximum profit from a contiguous period of stock price changes without exceeding a risk threshold. Which data structure would you use to support real‑time queries?

A balanced binary search tree (e.g., TreeSet in Java or std::set in C++) or a Fenwick tree with coordinate compression can store prefix sums. They allow O(log n) insertion and query for the largest prefix ≤ currentPrefix - target, enabling real‑time updates as new price changes arrive.

Q3During a high‑growth startup interview, you are asked to explain why the O(n log n) solution is preferable over an O(n^2) brute force, even if the input size is moderate. What points would you highlight?

I would emphasize scalability: the quadratic algorithm quickly becomes a bottleneck as data grows, leading to unacceptable latency. The O(n log n) solution offers a predictable performance curve, uses only linear additional space, and can be further optimized with a two‑pointer approach if constraints allow. Demonstrating awareness of algorithmic complexity shows readiness for production‑grade systems.

Examples

Example 1

Input

[4, 5, 3, 4, 5]

Output

Explanation: Step-by-step: Given the array [4, 5, 3, 4, 5], we can select the subarray [4, 5] and adjust it to 9 by increasing the weight of the first crate by 4 units and decreasing the weight of the third crate by 4 units.

Example 2

Input

[3, -4, 5, 4, 5]

Output

Explanation: Step-by-step: Given the array [3, -4, 5, 4, 5], we can select the subarray [3, -4, 5] and adjust it to 8 by increasing the weight of the first crate by 4 units and decreasing the weight of the third crate by 4 units.

Constraints

1 <= weights.length <= 1000
-1000 <= weights[i] <= 1000
1 <= adjustments <= 8
adjustments <= weights.length

Optimal Approach & Strategy

Compute prefix sums and maintain a sorted set of seen prefixes. For each prefix, binary‑search the largest earlier prefix ≤ currentPrefix - target to maximize the subarray sum, achieving O(n log n) time.

Brute Force Approach

Enumerate all O(n^2) subarrays, compute each sum, and keep the maximum that does not exceed the target.

Code Solutions

Language:

JavaScript Solution

Time: O(n log n)

/**
 * @param {number[]} crates
 * @return {number}
 */
var maxSupplyWeight = function(crates) {
    if (crates.length === 0) return 0;
    
    let maxEndingHere = crates[0];
    let maxSoFar = crates[0];
    
    for (let i = 1; i < crates.length; i++) {
        maxEndingHere = Math.max(crates[i], maxEndingHere + crates[i]);
        maxSoFar = Math.max(maxSoFar, maxEndingHere);
    }
    
    return maxSoFar;
};

class Solution {
public:
    int maxSupplyWeight(vector<int>& crates) {
        int n = crates.size();
        if (n == 0) return 0;
        
        // Prefix sums
        vector<int> prefix(n + 1, 0);
        for (int i = 0; i < n; ++i) {
            prefix[i + 1] = prefix[i] + crates[i];
        }
        
        // We want to find max subarray sum.
        // Using Kadane's algorithm for O(n) time.
        int maxEndingHere = crates[0];
        int maxSoFar = crates[0];
        
        for (int i = 1; i < n; ++i) {
            maxEndingHere = max(crates[i], maxEndingHere + crates[i]);
            maxSoFar = max(maxSoFar, maxEndingHere);
        }
        
        return maxSoFar;
    }
};

class Solution {
    public int maxSupplyWeight(int[] crates) {
        if (crates.length == 0) return 0;
        
        int maxEndingHere = crates[0];
        int maxSoFar = crates[0];
        
        for (int i = 1; i < crates.length; i++) {
            maxEndingHere = Math.max(crates[i], maxEndingHere + crates[i]);
            maxSoFar = Math.max(maxSoFar, maxEndingHere);
        }
        
        return maxSoFar;
    }
}

class Solution:
    def maxSupplyWeight(self, crates: List[int]) -> int:
        if not crates:
            return 0
        
        max_ending_here = crates[0]
        max_so_far = crates[0]
        
        for i in range(1, len(crates)):
            max_ending_here = max(crates[i], max_ending_here + crates[i])
            max_so_far = max(max_so_far, max_ending_here)
        
        return max_so_far

/**
 * @param {number[]} crates
 * @return {number}
 */
var maxSupplyWeight = function(crates) {
    if (crates.length === 0) return 0;
    
    let maxEndingHere = crates[0];
    let maxSoFar = crates[0];
    
    for (let i = 1; i < crates.length; i++) {
        maxEndingHere = Math.max(crates[i], maxEndingHere + crates[i]);
        maxSoFar = Math.max(maxSoFar, maxEndingHere);
    }
    
    return maxSoFar;
};

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