Network Protocol Synthesizer 20 — Problem Statement & Solution Guide
Problem Description
In a distributed network monitoring system, a stream of integer telemetry values is received from various nodes. Each value represents a specific protocol metric. The system requires an aggregation routine that isolates and sums only those metrics that meet or exceed a defined threshold K. This operation is critical for filtering out noise and focusing on high-impact data points for subsequent analysis.
Given an array of integers metrics and an integer threshold K, your task is to compute the sum of all elements in metrics that are greater than or equal to K. If no elements satisfy this condition, the result should be 0. The solution must efficiently process the array to produce the correct aggregate value.
The input consists of a single array of integers and a single integer threshold. The output is a single integer representing the computed sum. Ensure that your implementation handles edge cases such as empty arrays or thresholds that exceed all metric values.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Protocol Synthesizer 20"
WHY DOES IT MATTER?
The filter‑and‑aggregate pattern is fundamental for real‑time analytics, enabling systems to extract meaningful signals from high‑velocity data without storing the entire dataset.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the sum can be updated incrementally, eliminating the need for auxiliary data structures or multiple passes, thus collapsing both time and space to their theoretical minima.
REAL-WORLD CONNECTION
Think of a network router that only forwards packets exceeding a certain priority level; it inspects each packet once and decides instantly, mirroring the one‑pass threshold check and accumulation.
During an interview, write the loop first, then add the conditional check; keep the code tight and avoid premature micro‑optimizations—clarity beats cleverness for this straightforward task.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a single-pass aggregation over a sequence of integers, a classic example of linear-time streaming algorithms. A naive solution might attempt to sort or store all values before processing, which inflates both time (O(N log N)) and space (O(N)) and becomes infeasible for massive telemetry streams. The optimal paradigm leverages the fact that the required operation—filtering by a threshold and summing—can be performed incrementally: as each value arrives, we compare it to K and, if it qualifies, add it to a running total. This approach embodies the "online" or "streaming" algorithmic pattern, guaranteeing O(N) time and O(1) auxiliary space regardless of input size.
Interview Questions on This Problem
Q1How would you compute the sum of all elements in an array that are greater than or equal to a given threshold K in a single pass?
Initialize a variable sum = 0, iterate through the array, and for each element x, if x >= K add x to sum. After the loop, sum holds the required total. This runs in O(N) time and O(1) extra space.
Q2Why is sorting the array before summing elements >= K not an optimal solution for this problem?
Sorting costs O(N log N) time and O(N) space (or O(1) in-place but still O(N log N) time). Since the sum can be computed without ordering, sorting adds unnecessary overhead and fails the time constraints for large N.
Q3In a distributed telemetry system, how can you compute the global sum of values >= K using map‑reduce?
Each mapper processes its partition, applying the same filter‑and‑sum logic locally, emitting a partial sum. The reducer then aggregates all partial sums into the final result. This preserves linear work per node and constant extra memory per mapper.
Examples
Input
metrics = [12, 5, 8, 15, 3], K = 10
Output
27
Explanation: Iterate through the array: 12 >= 10 (add 12), 5 < 10 (skip), 8 < 10 (skip), 15 >= 10 (add 15), 3 < 10 (skip). Sum = 12 + 15 = 27.
Input
metrics = [1, 2, 3, 4, 5], K = 10
Output
0
Explanation: No element in the array is greater than or equal to 10. Therefore, the sum remains 0.
Input
metrics = [10, 10, 10], K = 10
Output
30
Explanation: All elements are equal to the threshold K. 10 >= 10 (add 10), 10 >= 10 (add 10), 10 >= 10 (add 10). Sum = 10 + 10 + 10 = 30.
Input
metrics = [-5, -10, 0, 5], K = -5
Output
0
Explanation: Check each element against K = -5: -5 >= -5 (add -5), -10 < -5 (skip), 0 >= -5 (add 0), 5 >= -5 (add 5). Sum = -5 + 0 + 5 = 0.
Constraints
- 1 <= metrics.length <= 10^5
- -10^9 <= metrics[i] <= 10^9
- -10^9 <= K <= 10^9
Optimal Approach & Strategy
Traverse the array once, adding each element to the answer only if it meets the threshold, using a single accumulator variable.
Brute Force Approach
Store all numbers, sort them, then iterate from the first element >= K and sum the rest.
Verified Code Solutions
/**
* @param {number[]} metrics
* @param {number} K
* @return {number}
*/
var sumMetricsAboveThreshold = function(metrics, K) {
let total = 0;
for (let m of metrics) {
if (m >= K) {
total += m;
}
}
return total;
};#include <iostream>
#include <vector>
using namespace std;
class Solution {
public:
int sumMetricsAboveThreshold(vector<int>& metrics, int K) {
int total = 0;
for (int m : metrics) {
if (m >= K) {
total += m;
}
}
return total;
}
};import java.util.*;
class Solution {
public int sumMetricsAboveThreshold(int[] metrics, int K) {
int total = 0;
for (int m : metrics) {
if (m >= K) {
total += m;
}
}
return total;
}
}from typing import List
class Solution:
def sumMetricsAboveThreshold(self, metrics: List[int], K: int) -> int:
total = 0
for m in metrics:
if m >= K:
total += m
return total/**
* @param {number[]} metrics
* @param {number} K
* @return {number}
*/
var sumMetricsAboveThreshold = function(metrics, K) {
let total = 0;
for (let m of metrics) {
if (m >= K) {
total += m;
}
}
return total;
};Asked in Top Tech Interviews
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