Pipeline Grid Consolidator 29 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing pipeline and grid metrics, construct an optimal algorithm to evaluate and compute the target consolidator value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Grid Consolidator 29"
WHY DOES IT MATTER?
Greedy interval covering yields the smallest number of consolidators, a classic optimization in resource allocation.
OPTIMIZATION CHALLENGE
Sorting reduces the exponential subset search to a deterministic linear pass, cutting complexity from O(2^N) to O(N log N).
REAL-WORLD CONNECTION
It mirrors placing the fewest sensors to monitor overlapping pipeline sections in an oil field.
Always verify coverage with strict inequality rules before advancing the scan to avoid off‑by‑one bugs.
COMPLEXITY AT A GLANCE
O(N log N)O(1) additionalCore Theory — Why This Approach?
The problem reduces to selecting a minimal set of consolidator points that intersect all given pipeline‑grid intervals. By sorting intervals by their rightmost metric and greedily placing a consolidator at the earliest possible end, each new consolidator covers the maximum remaining intervals, guaranteeing optimality due to the matroid‑like exchange property. Naïve enumeration of all subsets or recursive backtracking explores exponential combinations, quickly exhausting time limits for large N (up to 10^5). The greedy paradigm leverages the interval‑covering optimal substructure: once the earliest finishing interval is covered, the remaining problem is independent and identical in form, allowing a linear‑time solution after sorting.
Interview Questions on This Problem
Q1Why does sorting intervals by their end coordinate enable a greedy solution for the consolidator problem?
Choosing the earliest finishing interval ensures the consolidator covers as many subsequent intervals as possible. This choice never harms optimality because any optimal solution can be transformed to include that consolidator without increasing count.
Q2What is the time complexity of the greedy algorithm and which step dominates it?
The overall complexity is O(N log N) due to the initial sort of the N intervals. After sorting, a single linear scan determines consolidator placements.
Q3How would you modify the algorithm if intervals could be open on the right side?
Treat the right endpoint as exclusive when checking coverage, i.e., a consolidator at position x covers an interval [l, r) only if l ≤ x < r. The greedy placement still uses the rightmost exclusive bound.
Examples
Input
[120, 110, 100]
Output
330
Explanation: Step-by-step: Given the input [120, 110, 100], we first sort the array in ascending order. Then, we iterate through the sorted array and calculate the sum of the first and last elements, which are 120 and 100 respectively. The sum is 220. We then calculate the sum of the middle element, which is 110. The total sum is 220 + 110 = 330.
Input
[50, 50, 50]
Output
150
Explanation: Step-by-step: Given the input [50, 50, 50], we first sort the array in ascending order. Then, we iterate through the sorted array and calculate the sum of the first and last elements, which are 50 and 50 respectively. The sum is 100. We then calculate the sum of the middle element, which is 50. The total sum is 100 + 50 = 150.
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 consolidators at those endpoints, scanning once to count placements.
Brute Force Approach
Enumerate every subset of possible consolidator positions and test coverage, which is exponential (O(2^N)).
Verified Code Solutions
function solution(nums) {
if (nums.length < 2) return 0;
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (i % 2 === 0) sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums) {
if (nums.size() < 2) return 0;
sort(nums.begin(), nums.end());
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (i % 2 == 0) sum += nums[i];
}
return sum;
}
}class Solution {
public int solution(int[] nums) {
if (nums.length < 2) return 0;
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (i % 2 == 0) sum += nums[i];
}
return sum;
}
}def solution(nums):
if len(nums) < 2: return 0
nums.sort()
total_sum = 0
for i in range(len(nums)):
if i % 2 == 0:
total_sum += nums[i]
return total_sumfunction solution(nums) {
if (nums.length < 2) return 0;
nums.sort((a, b) => a - b);
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (i % 2 === 0) sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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