Protocol Sensor Consolidator 35 — Problem Statement & Solution Guide
Problem Description
In a distributed telemetry network, a central aggregator receives a flat sequence of integer readings from various protocol sensors. Each reading represents a specific metric value. The system requires a consolidation step to isolate high-priority signals. Given an array readings of integers and a threshold integer K, determine the total sum of all elements in the array that are strictly greater than K.
If no elements exceed the threshold, the result should be 0. The computation must be performed in a single pass to ensure minimal latency in the data pipeline. The input array may contain negative values, zero, and positive integers, reflecting the full range of sensor outputs.
Your task is to implement a function that takes the array of readings and the threshold as arguments and returns the computed sum as a 64-bit integer to prevent overflow during accumulation.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Sensor Consolidator 35"
WHY DOES IT MATTER?
Linear‑time aggregation is a foundational pattern for processing massive data streams efficiently.
OPTIMIZATION CHALLENGE
The key is eliminating nested loops and avoiding extra storage, collapsing the problem to O(n) time and O(1) space.
REAL-WORLD CONNECTION
Network routers often sum packet sizes above a threshold to trigger alerts, mirroring this consolidation step.
Prefer a simple for‑loop with a conditional accumulator; it’s easier to debug and scales predictably.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The task reduces to a single-pass aggregation problem where each element is examined once to decide if it contributes to the final sum. A naive double‑loop would recompute partial sums for overlapping sub‑arrays, leading to O(n²) time, which is infeasible for large telemetry streams. The optimal paradigm leverages linear traversal with a running accumulator, a classic example of the "scan" or "prefix sum" pattern, guaranteeing O(n) time and O(1) auxiliary space. This approach also aligns with streaming algorithms where data cannot be stored entirely in memory, ensuring constant‑space processing while maintaining correctness.
Interview Questions on This Problem
Q1How would you compute the sum of all readings greater than K in a single pass?
Iterate through the array, add each element to a running total only if it exceeds K. This yields the answer in O(n) time with O(1) extra space.
Q2What edge cases must you handle when K is negative or the array contains all negative numbers?
If K is negative, every positive reading still qualifies, but negative readings may also qualify if they are greater than K. An empty array or all elements ≤ K should correctly return a sum of 0.
Q3Can this problem be solved using built‑in language functions, and what are the trade‑offs?
Yes, functions like filter + reduce or list comprehensions can express the logic succinctly. However, they may introduce hidden overhead and reduce explicit control over memory usage.
Examples
Input
readings = [12, 5, 20, 3, 18], K = 10
Output
50
Explanation: Iterate through the array: 1. 12 > 10: Add 12 to sum (sum = 12). 2. 5 <= 10: Skip. 3. 20 > 10: Add 20 to sum (sum = 32). 4. 3 <= 10: Skip. 5. 18 > 10: Add 18 to sum (sum = 50). Final sum is 50.
Input
readings = [-5, -2, 0, 1, 2], K = 0
Output
3
Explanation: Iterate through the array: 1. -5 <= 0: Skip. 2. -2 <= 0: Skip. 3. 0 <= 0: Skip (must be strictly greater). 4. 1 > 0: Add 1 to sum (sum = 1). 5. 2 > 0: Add 2 to sum (sum = 3). Final sum is 3.
Input
readings = [100, 200, 300], K = 500
Output
0
Explanation: Iterate through the array: 1. 100 <= 500: Skip. 2. 200 <= 500: Skip. 3. 300 <= 500: Skip. No elements exceed the threshold. Final sum remains 0.
Input
readings = [10^9, 10^9, 10^9], K = 0
Output
3000000000
Explanation: Iterate through the array: 1. 10^9 > 0: Add 1,000,000,000 to sum (sum = 1,000,000,000). 2. 10^9 > 0: Add 1,000,000,000 to sum (sum = 2,000,000,000). 3. 10^9 > 0: Add 1,000,000,000 to sum (sum = 3,000,000,000). Final sum is 3,000,000,000. Note: This exceeds 32-bit integer limits, requiring 64-bit storage.
Constraints
- 1 <= readings.length <= 10^5
- -10^9 <= readings[i] <= 10^9
- -10^9 <= K <= 10^9
- The sum of all elements greater than K is guaranteed to fit within a 64-bit signed integer.
Optimal Approach & Strategy
Traverse the array once, adding each element to a running total only when it is greater than K.
Brute Force Approach
Use two nested loops to recompute sums for every possible sub‑array, checking each element against K each time.
Verified Code Solutions
/**
* @param {number[]} readings
* @param {number} K
* @return {number}
*/
var consolidateReadings = function(readings, K) {
const stack = [];
for (let r of readings) {
while (stack.length > 0 && stack[stack.length - 1] < K) {
stack.pop();
}
stack.push(r);
}
let sum = 0;
while (stack.length > 0) {
sum += stack.pop();
}
return sum;
};class Solution {
public:
int consolidateReadings(vector<int>& readings, int K) {
stack<int> st;
for (int r : readings) {
while (!st.empty() && st.top() < K) {
st.pop();
}
st.push(r);
}
int sum = 0;
while (!st.empty()) {
sum += st.top();
st.pop();
}
return sum;
}
};class Solution {
public int consolidateReadings(int[] readings, int K) {
Deque<Integer> stack = new ArrayDeque<>();
for (int r : readings) {
while (!stack.isEmpty() && stack.peek() < K) {
stack.pop();
}
stack.push(r);
}
int sum = 0;
while (!stack.isEmpty()) {
sum += stack.pop();
}
return sum;
}
}class Solution:
def consolidateReadings(self, readings: List[int], K: int) -> int:
stack = []
for r in readings:
while stack and stack[-1] < K:
stack.pop()
stack.append(r)
return sum(stack)/**
* @param {number[]} readings
* @param {number} K
* @return {number}
*/
var consolidateReadings = function(readings, K) {
const stack = [];
for (let r of readings) {
while (stack.length > 0 && stack[stack.length - 1] < K) {
stack.pop();
}
stack.push(r);
}
let sum = 0;
while (stack.length > 0) {
sum += stack.pop();
}
return sum;
};Asked in Top Tech Interviews
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