Pipeline Vector Detector 47 — Problem Statement & Solution Guide
Problem Description
You are tasked with implementing a signal processing routine for a distributed sensor network. The network transmits a sequence of integer readings, and you must compute the 'detector value' based on a specific threshold. The detector value is defined as the sum of all readings in the sequence that are strictly greater than the given threshold k.
Given an array of integers representing the sensor readings and an integer threshold k, return the sum of all elements in the array that exceed k. If no elements exceed the threshold, return 0.
This problem requires an efficient linear scan through the data to aggregate the relevant values, ensuring optimal performance for large datasets.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Pipeline Vector Detector 47"
WHY DOES IT MATTER?
Efficient range aggregation turns repeated linear scans into fast lookups, crucial for high‑throughput systems.
OPTIMIZATION CHALLENGE
The key is reducing per‑query work from O(n) to O(log n) by preprocessing sorted data with suffix sums.
REAL-WORLD CONNECTION
Think of a sensor hub that must constantly report totals above dynamic safety thresholds.
Always profile query patterns first; if thresholds change often, invest in preprocessing, otherwise a simple scan suffices.
COMPLEXITY AT A GLANCE
O(n log n + Q log n)O(n)Core Theory — Why This Approach?
The naive solution iterates through the entire array, adding each element that exceeds k. While this runs in O(n) time, it becomes inefficient when the same dataset must answer many threshold queries, because each query repeats the full scan. An optimal paradigm preprocesses the data—by sorting the array and building a suffix sum array—so each query can be answered in O(log n) via binary search, reducing total work for Q queries to O(n log n + Q log n). This approach leverages the divide‑and‑conquer principle and prefix‑sum technique, turning repeated linear scans into logarithmic lookups.
Interview Questions on This Problem
Q1How would you handle multiple threshold queries efficiently?
Sort the array once and compute a suffix sum array; each query then uses binary search to find the first element > k and reads the pre‑computed sum. This reduces per‑query time to O(log n).
Q2What is the time‑space trade‑off when using a suffix sum array?
You gain O(1) query time after O(log n) binary search at the cost of O(n) extra space for the suffix sums. The preprocessing cost is O(n log n) for sorting plus O(n) for building the sums.
Q3Why might a Trie be unsuitable for this problem?
A Trie excels at prefix‑based lookups on strings or bit patterns, not at numeric range aggregation. Using it here adds unnecessary complexity without improving query performance.
Examples
Input
readings = [12, 5, 23, 8, 34], k = 10
Output
57
Explanation: 1. Initialize sum = 0. 2. Check 12: 12 > 10, so sum = 0 + 12 = 12. 3. Check 5: 5 <= 10, skip. 4. Check 23: 23 > 10, so sum = 12 + 23 = 35. 5. Check 8: 8 <= 10, skip. 6. Check 34: 34 > 10, so sum = 35 + 34 = 69. 7. Return 69.
Input
readings = [1, 2, 3, 4, 5], k = 10
Output
0
Explanation: 1. Initialize sum = 0. 2. Check 1: 1 <= 10, skip. 3. Check 2: 2 <= 10, skip. 4. Check 3: 3 <= 10, skip. 5. Check 4: 4 <= 10, skip. 6. Check 5: 5 <= 10, skip. 7. No elements exceed k, so return 0.
Input
readings = [100, 200, 300], k = 150
Output
500
Explanation: 1. Initialize sum = 0. 2. Check 100: 100 <= 150, skip. 3. Check 200: 200 > 150, so sum = 0 + 200 = 200. 4. Check 300: 300 > 150, so sum = 200 + 300 = 500. 5. Return 500.
Input
readings = [-5, -10, 0, 5, 10], k = -1
Output
15
Explanation: 1. Initialize sum = 0. 2. Check -5: -5 <= -1, skip. 3. Check -10: -10 <= -1, skip. 4. Check 0: 0 > -1, so sum = 0 + 0 = 0. 5. Check 5: 5 > -1, so sum = 0 + 5 = 5. 6. Check 10: 10 > -1, so sum = 5 + 10 = 15. 7. Return 15.
Constraints
- 1 <= readings.length <= 10^5
- -10^9 <= readings[i] <= 10^9
- -10^9 <= k <= 10^9
Optimal Approach & Strategy
Sort once, compute suffix sums, then answer each query with binary search in O(log n).
Brute Force Approach
Iterate through the list and add each element that is > k; O(n) per query.
Verified Code Solutions
/**
* @param {number[]} readings
* @param {number} k
* @return {number}
*/
var detectorValue = function(readings, k) {
return readings.reduce((acc, r) => r > k ? acc + r : acc, 0);
};class Solution {
public:
int detectorValue(vector<int>& readings, int k) {
int sum = 0;
for (int r : readings) {
if (r > k) {
sum += r;
}
}
return sum;
}
};class Solution {
public int detectorValue(int[] readings, int k) {
int sum = 0;
for (int r : readings) {
if (r > k) {
sum += r;
}
}
return sum;
}
}class Solution:
def detectorValue(self, readings: List[int], k: int) -> int:
return sum(r for r in readings if r > k)/**
* @param {number[]} readings
* @param {number} k
* @return {number}
*/
var detectorValue = function(readings, k) {
return readings.reduce((acc, r) => r > k ? acc + r : acc, 0);
};Asked in Top Tech Interviews
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