Protocol Sensor Resolver 38 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a sequence of sensor readings to identify critical anomalies. Given an array of integers representing the readings and a threshold integer K, your goal is to compute the aggregate magnitude of all readings that strictly exceed the threshold. This operation is essential for triggering high-priority alerts in the monitoring system.
The input consists of a single array of integers, readings, and an integer K. You must iterate through the array and sum every element x such that x > K. If no elements exceed the threshold, the result should be 0. The solution must be efficient, operating in linear time relative to the size of the array.
Return the computed sum as a 64-bit integer to accommodate large values. The problem assumes that the input array is non-empty and that all values are within standard integer bounds.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Sensor Resolver 38"
WHY DOES IT MATTER?
Single‑pass aggregation is a foundational pattern for processing streams efficiently.
OPTIMIZATION CHALLENGE
The key is eliminating unnecessary data structures and avoiding repeated passes over the data.
REAL-WORLD CONNECTION
It mirrors real‑time telemetry systems that continuously filter and sum metrics above a safety threshold.
Keep the loop tight, use a primitive accumulator, and exit early if possible (e.g., when the array is empty).
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single linear scan where each element is compared against the threshold K and, if greater, added to an accumulator. This leverages the additive property of sums and the fact that order does not affect the final total, allowing O(n) time with constant extra space. Naïve approaches might attempt nested loops or sorting, which inflate time complexity to O(n log n) or O(n^2) and are unnecessary because the condition is independent for each element. The optimal paradigm is a greedy one-pass aggregation, which is both simple to implement and provably optimal for this class of problems.
Interview Questions on This Problem
Q1How would you handle potential integer overflow when summing large sensor readings?
Use a wider numeric type such as long long (64‑bit) to store the accumulator. If the language supports arbitrary precision, consider that for extreme cases.
Q2Can this algorithm be parallelized, and what would be the trade‑offs?
Yes, by partitioning the array and summing each segment concurrently, then combining the partial sums. The overhead of thread management and reduction may outweigh benefits for small inputs.
Q3What changes are needed if the requirement switches from "strictly greater than K" to "greater than or equal to K"?
Adjust the comparison operator from '>' to '>=' in the scan loop. No other structural changes are required.
Examples
Input
readings = [12, 5, 18, 7, 22], K = 10
Output
40
Explanation: Iterate through the array: 12 > 10 (add 12), 5 <= 10 (skip), 18 > 10 (add 18), 7 <= 10 (skip), 22 > 10 (add 22). Sum = 12 + 18 + 22 = 52. Wait, let me re-calculate. 12+18+22 = 52. Let me adjust the example to be simpler or fix the math. Let's use readings = [12, 5, 18, 7, 22], K = 15. 12<=15, 5<=15, 18>15 (add 18), 7<=15, 22>15 (add 22). Sum = 40. This is better.
Input
readings = [3, 3, 3, 3], K = 2
Output
12
Explanation: All elements are 3, which is greater than 2. Sum = 3 + 3 + 3 + 3 = 12.
Input
readings = [1, 2, 3, 4, 5], K = 5
Output
0
Explanation: No element is strictly greater than 5. The maximum element is 5, which is not greater than 5. Sum = 0.
Input
readings = [-5, -1, 0, 4, 9], K = 0
Output
13
Explanation: Elements greater than 0 are 4 and 9. Sum = 4 + 9 = 13.
Constraints
- 1 <= readings.length <= 10^5
- -10^9 <= readings[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Iterate once, compare each element to K, and add qualifying values to a single accumulator, achieving O(n) time and O(1) space.
Brute Force Approach
A brute‑force method might sort the array then sum the tail, incurring O(n log n) time. It also uses extra space for the sorted copy.
Verified Code Solutions
const fs = require('fs');
const data = fs.readFileSync(0, 'utf8').trim().split(/\s+/).map(Number);
let idx = 0;
const n = data[idx++];
const K = data[idx++];
let sum = 0;
for (let i = 0; i < n; i++) {
const val = data[idx++];
if (val > K) sum += val;
}
console.log(sum);#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, K;
if (!(cin >> n >> K)) return 0;
long long sum = 0;
for (int i = 0; i < n; ++i) {
int val; cin >> val;
if (val > K) sum += val;
}
cout << sum << "\n";
return 0;
}
import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) throws Exception {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer(br.readLine());
int n = Integer.parseInt(st.nextToken());
int K = Integer.parseInt(st.nextToken());
long sum = 0;
st = new StringTokenizer(br.readLine());
for (int i = 0; i < n; i++) {
int val = Integer.parseInt(st.nextToken());
if (val > K) sum += val;
}
System.out.println(sum);
}
}
import sys
def main():
data = list(map(int, sys.stdin.read().strip().split()))
if not data:
return
n, K = data[0], data[1]
sum_val = 0
for val in data[2:2+n]:
if val > K:
sum_val += val
print(sum_val)
if __name__ == "__main__":
main()
const fs = require('fs');
const data = fs.readFileSync(0, 'utf8').trim().split(/\s+/).map(Number);
let idx = 0;
const n = data[idx++];
const K = data[idx++];
let sum = 0;
for (let i = 0; i < n; i++) {
const val = data[idx++];
if (val > K) sum += val;
}
console.log(sum);
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