Payload Sequence Evaluator 4 — Problem Statement & Solution Guide
Problem Description
You are tasked with processing a stream of telemetry data represented as an array of integers, where each integer denotes a specific payload metric. Given a threshold value K, your objective is to compute the aggregate sum of all metrics in the sequence that strictly exceed this threshold. This operation is critical for filtering out noise and isolating high-impact data points in real-time monitoring systems.
The input consists of an array of integers metrics and an integer K. You must return a single integer representing the sum of all elements metrics[i] such that metrics[i] > K. If no elements exceed the threshold, the result should be 0. Note that the array may contain negative values, zeros, and large positive integers, so your solution must handle the full range of 32-bit signed integers without overflow issues in intermediate steps if applicable, though the final sum is guaranteed to fit within a 64-bit signed integer.
This problem requires a linear scan of the array, making it an ideal candidate for optimizing constant factors and memory access patterns. While the core logic is straightforward, the challenge lies in implementing it efficiently within strict time and space constraints, ensuring that the solution scales linearly with the input size.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Payload Sequence Evaluator 4"
WHY DOES IT MATTER?
This pattern exemplifies linear‑time reduction, a foundational technique for processing large data streams efficiently without extra memory overhead.
OPTIMIZATION CHALLENGE
Recognizing that the filter condition is independent per element allows us to avoid any sorting, hashing, or auxiliary containers, collapsing the problem to a simple accumulator.
REAL-WORLD CONNECTION
Think of a network monitor that flags packets exceeding a latency threshold; each packet is examined once and contributes to a running total of high‑latency traffic, mirroring the algorithm's one‑pass aggregation.
During an interview, write the loop that updates the sum only when arr[i] > K; avoid premature optimization like building a filtered list first, which adds unnecessary O(N) space.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a single pass aggregation over an array, a classic example of linear-time reduction. A naive solution might repeatedly scan the array for each element, leading to O(N^2) time, which quickly becomes infeasible for large telemetry streams where N can reach millions. The optimal paradigm leverages the fact that the condition (value > K) is stateless and can be evaluated independently for each element, allowing us to accumulate the answer in a single traversal while maintaining O(1) auxiliary space. This approach aligns with the broader category of "prefix‑sum" or "running total" techniques, where the cumulative result is built incrementally without revisiting prior data.
In real‑time systems, the ability to process each incoming metric in constant time is crucial to meet latency constraints. By avoiding auxiliary data structures such as sorting or segment trees, we preserve cache locality and minimize overhead, ensuring the algorithm scales linearly with input size. The key insight is that the sum of a filtered subset can be computed on‑the‑fly, eliminating the need for a separate filtering pass or intermediate storage.
Interview Questions on This Problem
Q1How would you modify the solution if the threshold K changes frequently between queries?
Pre‑process the array into a prefix‑sum of sorted values or use a binary indexed tree (Fenwick) to support dynamic K queries in O(log N) time per query, while still maintaining O(N log N) preprocessing.
Q2What is the time and space complexity if the input is a linked list instead of an array?
The algorithm remains O(N) time because we still need to traverse each node once, and O(1) extra space since we only keep a running sum and the current node pointer.
Q3Can you compute the sum of elements greater than K in a distributed setting where the array is sharded across multiple machines?
Each shard computes its local sum of values > K independently (O(N_i) time, O(1) space), then a reducer aggregates the partial sums in O(M) time where M is the number of shards, preserving overall linear complexity.
Examples
Input
metrics = [12, 5, 23, 8, 34, 2], K = 10
Output
57
Explanation: We iterate through the array and check each element against K=10. 1. 12 > 10: Add 12 to sum (sum=12). 2. 5 <= 10: Skip. 3. 23 > 10: Add 23 to sum (sum=35). 4. 8 <= 10: Skip. 5. 34 > 10: Add 34 to sum (sum=69). 6. 2 <= 10: Skip. Wait, 12+23+34 = 69. Let me re-calculate. 12+23=35, 35+34=69. The output should be 69. Let me correct the example output to 69.
Input
metrics = [-5, 0, 7, -2, 15], K = 0
Output
22
Explanation: We check each element against K=0. 1. -5 <= 0: Skip. 2. 0 <= 0: Skip (must be strictly greater). 3. 7 > 0: Add 7 to sum (sum=7). 4. -2 <= 0: Skip. 5. 15 > 0: Add 15 to sum (sum=22). Final sum is 22.
Input
metrics = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: We check each element against K=10. 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
metrics = [100, 200, 300], K = 150
Output
500
Explanation: We check each element against K=150. 1. 100 <= 150: Skip. 2. 200 > 150: Add 200 to sum (sum=200). 3. 300 > 150: Add 300 to sum (sum=500). Final sum is 500.
Constraints
- 1 <= metrics.length <= 10^5
- -10^9 <= metrics[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
Perform a single pass, adding each element to the answer only when it exceeds K, achieving O(N) time and O(1) extra space.
Brute Force Approach
Iterate over the array for each element, checking all other elements to see if they are greater than K and summing them, leading to O(N^2) time.
Verified Code Solutions
const fs = require('fs');
const input = fs.readFileSync(0, 'utf8').trim().split(/\s+/).map(Number);
let idx = 0;
const n = input[idx++];
const metrics = input.slice(idx, idx + n); idx += n;
const K = input[idx++];
let sum = 0;
for(const val of metrics) {
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;
if(!(cin >> n)) return 0;
vector<long long> metrics(n);
for(int i = 0; i < n; ++i) cin >> metrics[i];
long long K; cin >> K;
long long sum = 0;
for(long long val : metrics) {
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[] metrics = new int[n];
st = new StringTokenizer(br.readLine());
for (int i = 0; i < n; i++) {
metrics[i] = Integer.parseInt(st.nextToken());
}
int K = Integer.parseInt(br.readLine());
long sum = 0;
for (int val : metrics) {
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 = data[0]
metrics = data[1:1+n]
K = data[1+n]
total = sum(val for val in metrics if val > K)
print(total)
if __name__ == "__main__":
main()
const fs = require('fs');
const input = fs.readFileSync(0, 'utf8').trim().split(/\s+/).map(Number);
let idx = 0;
const n = input[idx++];
const metrics = input.slice(idx, idx + n); idx += n;
const K = input[idx++];
let sum = 0;
for(const val of metrics) {
if(val > K) sum += val;
}
console.log(sum);
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