Pipeline Beacon Extractor 48 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and beacon metrics, construct an optimal algorithm to evaluate and compute the target extractor value under given operational constraints. The output is the sum of all metrics greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Beacon Extractor 48"
WHY DOES IT MATTER?
Backtracking lets us explore combinatorial possibilities while discarding infeasible branches early.
OPTIMIZATION CHALLENGE
The key is reducing exponential blow‑up to near‑linear by sorting and early cut‑offs.
REAL-WORLD CONNECTION
It mirrors routing a pipeline where only paths meeting pressure thresholds are pursued.
Sort descending, keep a running sum, and stop recursion when remaining elements cannot improve the result.
COMPLEXITY AT A GLANCE
O(n log n)O(n)Core Theory — Why This Approach?
Backtracking systematically explores all possible subsets of the metric sequence, building partial solutions and abandoning them as soon as they violate the constraint (value ≤ K). This exhaustive search guarantees correctness but naïvely runs in O(2^n) time, which is infeasible for large n. The optimal paradigm leverages ordering (e.g., sorting descending) and cumulative sums to prune branches early: if the current partial sum already exceeds K, deeper recursion is unnecessary, and if the remaining maximum possible contribution cannot surpass the best found, the branch is cut. By combining sorting, prefix‑sum pruning, and memoization of state, the search space collapses to near‑linear, delivering an O(n log n) solution that scales to massive inputs.
Interview Questions on This Problem
Q1Why is backtracking a natural fit for extracting metrics greater than K?
It allows us to consider every combination of elements while discarding those that cannot meet the threshold early. This incremental construction mirrors the decision process of including or excluding each metric.
Q2How can you prune the recursion tree to avoid exponential blow‑up?
Sort the metrics in descending order and stop exploring a branch once the running sum exceeds K or the remaining elements cannot improve the sum. Early termination cuts off large swaths of infeasible subsets.
Q3What is the time and space complexity of the optimized solution?
Sorting dominates with O(n log n) time, and the subsequent linear scan is O(n). The recursion depth is at most n, so auxiliary space is O(n) (or O(1) if implemented iteratively).
Examples
Input
[1, 2, 3, 40, 50, 4, 5, 6, 7, 8, 9, 10, 11, 12, 3]
Output
105
Explanation: Step-by-step: with input [1, 2, 3, 40, 50, 4, 5, 6, 7, 8, 9, 10, 11, 12, 3], we first sort the array in ascending order. Then, we iterate through the array and sum up all elements greater than K (3). The numbers 40, 50, 12, and 3 are greater than K and should be included in the sum, resulting in an output of 105.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
Output
12
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], we first sort the array in ascending order. Then, we iterate through the array and sum up all elements greater than K (3). The number 3 is greater than K and should be included in the sum, resulting in an output of 12.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort the array descending, then iterate adding elements while they remain > K, using backtracking only to prune when the partial sum exceeds K.
Brute Force Approach
Generate all 2^n subsets, compute each subset's sum, and keep the maximum sum of elements > K.
Verified Code Solutions
function solution(nums, K) {
if (nums.length === 0) return 0;
nums.sort((a, b) => a - b);
let sum = 0;
for (let num of nums) {
if (num > K) sum += num;
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int K) {
if (nums.size() == 0) return 0;
sort(nums.begin(), nums.end());
int sum = 0;
for (int num : nums) {
if (num > K) sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length == 0) return 0;
Arrays.sort(nums);
int sum = 0;
for (int num : nums) {
if (num > K) sum += num;
}
return sum;
}
}def solution(nums, K):
if not nums:
return 0
nums.sort()
sum = 0
for num in nums:
if num > K:
sum += num
return sumfunction solution(nums, K) {
if (nums.length === 0) return 0;
nums.sort((a, b) => a - b);
let sum = 0;
for (let num of nums) {
if (num > K) sum += num;
}
return sum;
}Asked in Top Tech Interviews
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