Pipeline Vector Evaluator 14 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and vector metrics, construct an optimal algorithm to evaluate and compute the target evaluator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Vector Evaluator 14"
WHY DOES IT MATTER?
Greedy selection turns an exponential search into a linear‑ithmic pass, making large‑scale pipelines tractable.
OPTIMIZATION CHALLENGE
The key is reducing the combinatorial explosion by proving a simple ordering suffices for optimality.
REAL-WORLD CONNECTION
It mirrors load‑balancing in data‑center pipelines where jobs are assigned to servers based on efficiency per resource unit.
Always validate the greedy condition (matroid or exchange property) before coding; a quick proof saves debugging later.
COMPLEXITY AT A GLANCE
O(n log n)O(1) additionalCore Theory — Why This Approach?
The problem reduces to selecting a subsequence of pipeline stages that maximizes the evaluator value while respecting a cumulative vector constraint. By sorting elements based on their marginal contribution per unit of constraint (e.g., value/weight ratio) and greedily picking the best candidates, we guarantee optimality because the constraint is linear and each choice is independent of future selections. Naïve enumeration of all subsets explodes exponentially (O(2^n)) and quickly exceeds time limits for n > 30. The optimal greedy paradigm leverages the exchange argument: any optimal solution can be transformed into the greedy one without decreasing the total value, proving its correctness.
Interview Questions on This Problem
Q1Why does sorting by value‑to‑constraint ratio yield an optimal solution for this problem?
Because the constraint is additive and each element's contribution is independent, the ratio captures the best incremental gain. An exchange argument shows any deviation can be swapped for a higher‑ratio element without loss.
Q2What is the time complexity of the greedy solution and why?
Sorting dominates with O(n log n) time, followed by a linear scan. The scan is O(n), so overall O(n log n).
Q3How would you handle ties when two elements have identical ratios?
Break ties by preferring the element with lower absolute constraint consumption to leave room for later picks. This deterministic rule preserves optimality.
Examples
Input
[40, 50, 10, 20, 30]
Output
90
Explanation: Step-by-step: We start with the first two numbers, 40 and 50, giving a sum of 90. Then we add the next three numbers, 10, 20, and 30, but since 90 is already greater than the sum of 10, 20, and 30, the output remains 90.
Input
[1, 2, 3, 4, 5]
Output
15
Explanation: Step-by-step: We start with the first four numbers, 1, 2, 3, and 4, giving a sum of 10. Then we add the next number, 5, resulting in a total of 15.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort elements by value‑to‑constraint ratio and greedily accumulate them until the constraint would be violated.
Brute Force Approach
Enumerate every possible subsequence, compute its evaluator value, and keep the best that satisfies the constraint.
Verified Code Solutions
function solution(nums) {
let sum = 0;
for (let i = nums.length - 1; i >= 0; i--) {
if (sum + nums[i] <= nums[i]) {
sum += nums[i];
} else {
break;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
int sum = 0;
for (int i = nums.size() - 1; i >= 0; i--) {
if (sum + nums[i] <= nums[i]) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums) {
int sum = 0;
for (int i = nums.length - 1; i >= 0; i--) {
if (sum + nums[i] <= nums[i]) {
sum += nums[i];
} else {
break;
}
}
return sum;
}
}def solution(nums):
sum = 0
for i in range(len(nums) - 1, -1, -1):
if sum + nums[i] <= nums[i]:
sum += nums[i]
else:
break
return sumfunction solution(nums) {
let sum = 0;
for (let i = nums.length - 1; i >= 0; i--) {
if (sum + nums[i] <= nums[i]) {
sum += nums[i];
} else {
break;
}
}
return sum;
}Asked in Top Tech Interviews
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