Pipeline Beacon Extractor 13 — 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.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Beacon Extractor 13"
WHY DOES IT MATTER?
Binary search turns exponential‑time brute force into logarithmic time, making large‑scale queries feasible.
OPTIMIZATION CHALLENGE
The key is reducing the search interval by half each step while preserving the invariant that the answer lies within the bounds.
REAL-WORLD CONNECTION
It mirrors how a GPS narrows down a location by repeatedly halving the search area.
Always maintain a clear invariant and use safe mid computation to avoid overflow and off‑by‑one bugs.
COMPLEXITY AT A GLANCE
O(log n)O(1)Core Theory — Why This Approach?
Binary search exploits a monotonic predicate over a sorted or otherwise ordered domain, repeatedly halving the search interval to locate the boundary where the predicate flips. This divide‑and‑conquer approach reduces the number of inspections from linear to logarithmic, guaranteeing O(log n) time on large inputs. Naïve linear scans examine every element, leading to O(n) time which becomes prohibitive when n reaches millions or when the predicate evaluation itself is costly. The optimal paradigm frames the problem as finding the smallest index (or value) satisfying the condition and uses two‑pointer bounds that converge, ensuring both correctness and minimal comparisons.
Interview Questions on This Problem
Q1How does binary search guarantee O(log n) time complexity?
Each iteration discards half of the remaining search space, so the number of steps grows logarithmically with the input size. After at most ⌈log₂ n⌉ iterations the interval collapses to a single element.
Q2When can binary search be applied beyond sorted arrays?
It works on any monotonic predicate where the answer transitions from false to true (or vice‑versa) as the input increases. Examples include searching for a minimum feasible value in parametric problems.
Q3What are common pitfalls when implementing binary search on integer ranges?
Using mid = (low + high) / 2 can overflow; use mid = low + (high‑low)/2 instead. Also, forgetting to move the correct bound can cause infinite loops or off‑by‑one errors.
Examples
Input
[1, 2, 3, 4, 5], 3
Output
12
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5] and the target value 3, we first check if all elements in the array are less than 3. Since 1 and 2 are less than 3, we return 0. However, we should also check if all elements are greater than or equal to 3. Since 3, 4, and 5 are greater than or equal to 3, we should return the sum of these elements, which is 3 + 4 + 5 = 12.
Input
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200], 0
Output
100
Explanation: Step-by-step: Given the input array [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133, 134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159, 160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 186, 187, 188, 189, 190, 191, 192, 193, 194, 195, 196, 197, 198, 199, 200] and the target value 0, we first check if all elements in the array are less than 0. Since all elements are greater than or equal to 0, we return the sum of all elements in the array, which is 100.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Apply binary search on the implicit index space, adjusting low/high based on the predicate, achieving O(log n) time with O(1) extra space.
Brute Force Approach
Iterate through the sequence from start to finish, checking each element until the condition is met, which costs O(n) time.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] >= k) {
sum += nums[i];
}
}
return sum;
}class Solution {
public:
int solution(vector<int> nums, int k) {
int sum = 0;
for (int i = 0; i < nums.size(); i++) {
if (nums[i] >= k) {
sum += nums[i];
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
for (int i = 0; i < nums.length; i++) {
if (nums[i] >= k) {
sum += nums[i];
}
}
return sum;
}
}def solution(nums, k):
sum = 0
for i in range(len(nums)):
if nums[i] >= k:
sum += nums[i]
return sumfunction solution(nums, k) {
let sum = 0;
for (let i = 0; i < nums.length; i++) {
if (nums[i] >= k) {
sum += nums[i];
}
}
return sum;
}Asked in Top Tech Interviews
Solve in Interative Editor
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