Matrix Vessel Optimizer 7 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing matrix and vessel metrics, construct an optimal algorithm to evaluate and compute the target optimizer value under given operational constraints. The algorithm should select the first k elements from the sorted array in descending order, where k is the number of elements that can be selected to maximize the sum.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Matrix Vessel Optimizer 7"
WHY DOES IT MATTER?
The "select‑top‑k after sorting" pattern appears in many optimization scenarios—budget allocation, resource provisioning, and ranking systems—where you need the highest‑impact items under a simple additive objective. Mastering this pattern prevents over‑engineering and ensures you choose the mathematically optimal subset instantly.
OPTIMIZATION CHALLENGE
The key insight is that the sum function is monotonic with respect to element magnitude, allowing us to replace exhaustive search with a single sort or linear selection. This collapses the combinatorial explosion to O(n log n) or O(n) time, and O(1) additional space when sorting in‑place.
REAL-WORLD CONNECTION
Think of a data‑center load balancer that must pick the fastest servers to handle a burst of traffic. By ranking servers by latency (descending speed) and picking the top‑k that fit within the power budget, the balancer mirrors the same greedy selection logic used in this algorithm.
In an interview, sort the array first, then walk the sorted list while maintaining a running sum and a counter. Stop when the next element would violate the problem’s constraint. This one‑pass after sorting is both simple to code and provably optimal.
COMPLEXITY AT A GLANCE
O(n log n)O(1) additional (in‑place sort) or O(n) if using a separate arrayCore Theory — Why This Approach?
The core of this problem lies in the greedy selection principle: when the goal is to maximize the sum of a subset of elements under no additional constraints, the optimal subset is simply the collection of the largest values. A naive solution would enumerate all possible subsets, which grows exponentially (2^n) and quickly becomes infeasible for n > 30. By recognizing that the sum is a monotonic function of the chosen elements, we can sort the array in descending order and consider prefixes of this sorted list. The optimal k is the point where adding the next largest element no longer improves the objective—often this translates to taking all positive numbers or stopping at a given capacity constraint. This reduces the problem to a linear‑time selection (using QuickSelect) or an O(n log n) sort followed by a single pass, delivering a tractable solution for massive inputs.
Interview Questions on This Problem
Q1How would you compute the maximum possible sum of a subset when you can pick any number of elements from an unsorted array?
Sort the array in descending order (or use a max‑heap/QuickSelect) and then iterate from the largest element, accumulating the sum until a stopping condition is met (e.g., the next element is non‑positive or a capacity limit is reached). This greedy approach yields the optimal sum because any smaller element would only decrease the total.
Q2Explain why a brute‑force enumeration of all subsets is impractical for n = 10^5 and how the greedy method circumvents this.
Enumerating all subsets requires O(2^n) time, which is astronomically large for n = 10^5. The greedy method reduces the problem to sorting (O(n log n)) or linear selection (O(n)) and a single linear scan, turning an exponential problem into a polynomial one that easily fits within typical time limits.
Q3In a system where each element represents a vessel's throughput, how would you adapt the algorithm if a hard capacity C limits the total sum you can select?
After sorting descending, iterate and keep a running total. Add the next element only if total + element ≤ C. This maintains the greedy optimality because we always consider the largest remaining element that fits, analogous to the classic “knapsack with unlimited items but unit weight” where the greedy choice is optimal.
Examples
Input
[100, 90, 80, 70, 60, 50, 40, 30, 20, 10]
Output
340
Explanation: Step-by-step: Given the input array [100, 90, 80, 70, 60, 50, 40, 30, 20, 10], we first sort the array in descending order. Then, we select the first k elements from the sorted array, where k is the number of elements that can be selected to maximize the sum. In this case, k is equal to the length of the array, which is 10. The sum of the first k elements is 100 + 90 + 80 + 70 + 60 + 50 + 40 + 30 + 20 + 10 = 340.
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5], we first sort the array in descending order. Then, we select the first k elements from the sorted array, where k is the number of elements that can be selected to maximize the sum. In this case, k is equal to the length of the array, which is 5. The sum of the first k elements is 5 + 4 + 3 + 2 + 1 = 15.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort the array descending (or use QuickSelect) and take a prefix until the stopping condition is met, achieving O(n log n) time (or O(n) with selection).
Brute Force Approach
Enumerate every possible subset, compute its sum, and keep the maximum; this requires O(2^n) time.
Verified Code Solutions
function solution(nums, k) {
if (k > nums.length) {
k = nums.length;
}
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
if (k > nums.size()) {
k = nums.size();
}
sort(nums.begin(), nums.end(), greater<int>());
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
if (k > nums.length) {
k = nums.length;
}
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
if k > len(nums):
k = len(nums)
nums.sort(reverse=True)
sum = 0
for i in range(k):
sum += nums[i]
return sumfunction solution(nums, k) {
if (k > nums.length) {
k = nums.length;
}
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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