Sensor Cluster Architect 14 — Problem Statement & Solution Guide
Problem Description
In a distributed telemetry system, a central node receives a continuous stream of integer sensor readings. To optimize bandwidth, the system must filter this stream based on a dynamic threshold K. Your task is to compute the 'Architect Value', which is defined as the sum of all readings in the stream that are strictly greater than K.
Given an array of integers readings representing the incoming data and an integer K representing the threshold, return the sum of all elements x in readings such that x > K. If no elements exceed the threshold, return 0.
The solution must efficiently process the stream to determine the aggregate value of the high-priority signals.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Sensor Cluster Architect 14"
WHY DOES IT MATTER?
The single‑pass aggregation pattern is fundamental because it extracts required information from large datasets with minimal time and space, a common need in high‑throughput systems and interview problems.
OPTIMIZATION CHALLENGE
The key insight is recognizing that ordering is irrelevant for a sum‑based filter, allowing us to discard sorting or heap structures and achieve O(N) time with O(1) auxiliary space.
REAL-WORLD CONNECTION
In distributed telemetry, edge devices continuously emit metrics; a central aggregator must filter and sum values above a safety threshold without storing the entire history, mirroring this linear‑scan solution.
During an interview, first state the O(N) scan, then discuss why more complex structures like heaps or sorting are unnecessary, showing you can choose the simplest optimal solution.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem asks for the sum of all array elements that are strictly greater than a given threshold K. The most direct way to achieve this is to traverse the array once, checking each element against K and accumulating the qualifying values. This linear‑time solution leverages the fact that the operation required (a comparison and a conditional addition) is O(1) per element, leading to an overall O(N) time complexity where N is the number of readings.\n\nA naive approach might attempt to sort the array first and then sum the suffix that exceeds K. Sorting incurs O(N log N) time, which is unnecessary because the ordering of elements does not affect the sum. Moreover, sorting would also require additional space for the sorting algorithm (often O(N) for mergesort) or in‑place modifications that could be undesirable in a streaming or read‑only context.\n\nThe optimal paradigm therefore follows the “single‑pass aggregation” pattern: iterate, compare, and accumulate. This approach respects both time and space constraints, works seamlessly with streaming data, and aligns with the heap‑related theme by illustrating that not every heap problem needs a heap—recognizing when a simple scan suffices is a key skill.
Interview Questions on This Problem
Q1How would you compute the sum of all sensor readings greater than K in a single pass, and why is this preferable to sorting the array first?
Iterate through the array once, adding each element to a running total only if it exceeds K. This runs in O(N) time and O(1) extra space, whereas sorting would be O(N log N) time and potentially O(N) space, making the linear scan far more efficient for large streams.
Q2If the sensor readings were arriving as a real‑time stream rather than a static array, how could you adapt the solution while keeping memory usage minimal?
Maintain a running sum and, for each incoming reading, compare it to K and add it to the sum if it is larger. Since you only need the current sum and K, memory usage stays O(1) regardless of stream length.
Q3Explain a scenario where using a max‑heap would be beneficial for a similar problem, and why it would be overkill for the current task.
If the requirement were to retrieve the top‑M readings greater than K, a max‑heap of size M would allow O(N log M) insertion and O(M log M) extraction. For simply summing all values > K, a heap adds unnecessary log‑factor overhead; a linear scan is optimal.
Examples
Input
readings = [12, 5, 18, 3, 22], K = 10
Output
52
Explanation: Iterate through the readings: 1. 12 > 10, add 12 to sum (sum = 12). 2. 5 <= 10, skip. 3. 18 > 10, add 18 to sum (sum = 30). 4. 3 <= 10, skip. 5. 22 > 10, add 22 to sum (sum = 52). Final sum is 52.
Input
readings = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: Iterate through the readings: 1. 1 <= 10, skip. 2. 2 <= 10, skip. 3. 3 <= 10, skip. 4. 4 <= 10, skip. 5. 5 <= 10, skip. No elements exceed the threshold, so the sum remains 0.
Input
readings = [100, 200, 300], K = 50
Output
600
Explanation: Iterate through the readings: 1. 100 > 50, add 100 to sum (sum = 100). 2. 200 > 50, add 200 to sum (sum = 300). 3. 300 > 50, add 300 to sum (sum = 600). Final sum is 600.
Input
readings = [-5, -10, 0, 5, 10], K = 0
Output
15
Explanation: Iterate through the readings: 1. -5 <= 0, skip. 2. -10 <= 0, skip. 3. 0 <= 0, skip (must be strictly greater). 4. 5 > 0, add 5 to sum (sum = 5). 5. 10 > 0, add 10 to sum (sum = 15). Final sum is 15.
Constraints
- 1 <= readings.length <= 10^5
- -10^9 <= readings[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Traverse the array once, adding each element that exceeds K to a running total, achieving O(N) time and O(1) extra space.
Brute Force Approach
Sort the array and then sum the suffix of elements greater than K, which costs O(N log N) time.
Verified Code Solutions
/**
* @param {number[]} readings
* @param {number} K
* @return {number}
*/
var architectValue = function(readings, K) {
const heap = [];
let sum = 0;
const push = (val) => {
heap.push(val);
let i = heap.length - 1;
while (i > 0) {
const parent = Math.floor((i - 1) / 2);
if (heap[parent] > heap[i]) {
[heap[parent], heap[i]] = [heap[i], heap[parent]];
i = parent;
} else break;
}
};
const pop = () => {
const top = heap[0];
heap[0] = heap[heap.length - 1];
heap.pop();
let i = 0;
const n = heap.length;
while (true) {
let smallest = i;
const left = 2 * i + 1;
const right = 2 * i + 2;
if (left < n && heap[left] < heap[smallest]) smallest = left;
if (right < n && heap[right] < heap[smallest]) smallest = right;
if (smallest !== i) {
[heap[i], heap[smallest]] = [heap[smallest], heap[i]];
i = smallest;
} else break;
}
return top;
};
for (let r of readings) {
push(r);
sum += r;
if (heap.length > K) {
sum -= pop();
}
}
return sum;
};class Solution {
public:
int architectValue(vector<int>& readings, int K) {
priority_queue<int, vector<int>, greater<int>> minHeap;
int sum = 0;
for (int r : readings) {
minHeap.push(r);
sum += r;
if (minHeap.size() > K) {
sum -= minHeap.top();
minHeap.pop();
}
}
return sum;
}
};import java.util.PriorityQueue;
class Solution {
public int architectValue(int[] readings, int K) {
PriorityQueue<Integer> minHeap = new PriorityQueue<>();
int sum = 0;
for (int r : readings) {
minHeap.offer(r);
sum += r;
if (minHeap.size() > K) {
sum -= minHeap.poll();
}
}
return sum;
}
}import heapq
class Solution:
def architectValue(self, readings: List[int], K: int) -> int:
min_heap = []
total_sum = 0
for r in readings:
heapq.heappush(min_heap, r)
total_sum += r
if len(min_heap) > K:
total_sum -= heapq.heappop(min_heap)
return total_sum/**
* @param {number[]} readings
* @param {number} K
* @return {number}
*/
var architectValue = function(readings, K) {
const heap = [];
let sum = 0;
const push = (val) => {
heap.push(val);
let i = heap.length - 1;
while (i > 0) {
const parent = Math.floor((i - 1) / 2);
if (heap[parent] > heap[i]) {
[heap[parent], heap[i]] = [heap[i], heap[parent]];
i = parent;
} else break;
}
};
const pop = () => {
const top = heap[0];
heap[0] = heap[heap.length - 1];
heap.pop();
let i = 0;
const n = heap.length;
while (true) {
let smallest = i;
const left = 2 * i + 1;
const right = 2 * i + 2;
if (left < n && heap[left] < heap[smallest]) smallest = left;
if (right < n && heap[right] < heap[smallest]) smallest = right;
if (smallest !== i) {
[heap[i], heap[smallest]] = [heap[smallest], heap[i]];
i = smallest;
} else break;
}
return top;
};
for (let r of readings) {
push(r);
sum += r;
if (heap.length > K) {
sum -= pop();
}
}
return sum;
};Asked in Top Tech Interviews
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