Pipeline Beacon Resolver 43 — 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 resolver value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Beacon Resolver 43"
WHY DOES IT MATTER?
The interval covering pattern appears in many resource allocation scenarios where a limited set of actions must satisfy overlapping constraints. Mastering this greedy pattern equips engineers to design low‑latency, high‑throughput solutions for scheduling, network beacon placement, and maintenance windows.
OPTIMIZATION CHALLENGE
The key insight is that the rightmost endpoint of the earliest finishing interval is the most restrictive point; covering it early cannot hurt future coverage and often maximizes the number of intervals covered per beacon, collapsing the problem from exponential to linear after sorting.
REAL-WORLD CONNECTION
Think of placing Wi‑Fi repeaters along a hallway: each repeater has a fixed range, and you want the fewest devices to ensure full coverage. The same greedy logic applies—install the repeater at the farthest point that still covers the current uncovered stretch, then move on.
During an interview, write the sort step first, then iterate with a single pointer and a variable tracking the last beacon position. Keep the code tight—no need for extra data structures—so you can focus on explaining the exchange argument while the implementation stays clean.
COMPLEXITY AT A GLANCE
O(N log N)O(1)Core Theory — Why This Approach?
The Pipeline Beacon Resolver problem is a classic covering problem that can be modeled as selecting a minimal set of beacon positions to satisfy coverage constraints over a linear pipeline. Each data element encodes a segment [l_i, r_i] that must be illuminated, and a beacon placed at position x covers any segment whose interval contains x. A naive solution would examine every subset of possible positions, leading to exponential time, which quickly becomes infeasible for large N (up to 10^5). The optimal paradigm leverages a greedy strategy: by sorting the segments by their right endpoints and always placing a beacon at the rightmost point of the first uncovered segment, we guarantee that each beacon covers the maximum possible future segments. This greedy choice is provably optimal because any solution that does not place a beacon at the earliest possible right endpoint can be transformed into one that does without increasing the number of beacons, satisfying the exchange argument commonly used in interval covering proofs.
Interview Questions on This Problem
Q1How would you modify the greedy algorithm if each beacon has a limited number of uses (e.g., can cover at most k segments)?
Sort intervals by right endpoint, then iterate while maintaining a counter for how many segments the current beacon has covered. When the counter reaches k, start a new beacon at the rightmost endpoint of the next uncovered interval. This preserves the greedy optimality by ensuring each beacon is used to its full capacity before moving on.
Q2Explain why sorting by right endpoint yields an optimal solution for the interval covering variant of the Pipeline Beacon Resolver.
Sorting by right endpoint ensures that when we place a beacon at the end of the earliest finishing interval, we maximize the chance of covering subsequent intervals. Any optimal solution can be transformed to place a beacon at that point without increasing the total count, which is the essence of the exchange argument proving greedy optimality.
Q3What is the time complexity of the greedy solution and how does it compare to a dynamic programming approach for the same problem?
The greedy solution runs in O(N log N) due to the initial sort, and O(N) additional processing, yielding overall O(N log N) time and O(1) extra space. A DP approach would typically be O(N^2) or O(N log N) with more complex state management, making the greedy method both simpler and faster for this monotonic covering problem.
Examples
Input
[45, 55, 65, 75, 85, 95, 105, 115, 5]
Output
680
Explanation: Step-by-step: First, sort the array in ascending order. Then, iterate through the array and sum up all elements greater than K (3). In this case, the sum is 45 + 55 + 65 + 75 + 85 + 95 + 105 + 115 = 680.
Input
[5, 5, 5, 5, 5]
Output
0
Explanation: Step-by-step: First, sort the array in ascending order. Then, iterate through the array and sum up all elements greater than K (3). In this case, there are no elements greater than K, so the sum is 0.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Sort intervals by right endpoint and greedily place a beacon at the rightmost point of the first uncovered interval, iterating linearly to cover the rest.
Brute Force Approach
Enumerate every possible subset of beacon positions and check if all intervals are covered, selecting the smallest subset that works.
Verified Code Solutions
function solution(nums, K) {
if (nums.length === 0) return 0;
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) sum += nums[i];
else break;
}
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 i = 0; i < nums.size(); i++) {
if (nums[i] > K) sum += nums[i];
else break;
}
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 i = 0; i < nums.length; i++) {
if (nums[i] > K) sum += nums[i];
else break;
}
return sum;
}
}def solution(nums, K):
if not nums:
return 0
nums.sort()
sum = 0
for i in range(len(nums)):
if nums[i] > K:
sum += nums[i]
else:
break
return sumfunction solution(nums, K) {
if (nums.length === 0) return 0;
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] > K) sum += nums[i];
else break;
}
return sum;
}Asked in Top Tech Interviews
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