Network Node Detector 2 — Problem Statement & Solution Guide
Problem Description
Network Node Detector 2
You are given a list of integer metrics that represent the performance of various nodes in a network. Alongside this list, an integer threshold K is provided. Your task is to compute the sum of all metrics that are strictly greater than K. The algorithm should process the list efficiently, even for large inputs.
Input format:
- The first line contains an integer N, the number of metrics.
- The second line contains N space‑separated integers, the metrics themselves.
- The third line contains the integer K.
Output format:
- Output a single integer: the sum of all metrics that exceed K.
If no metric is greater than K, the sum is 0.
The problem can be solved using a simple linear scan, but you may also implement a recursive backtracking approach that visits each element and accumulates the sum when the condition is met.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Node Detector 2"
WHY DOES IT MATTER?
The linear‑scan (two‑pointer) pattern is essential for any problem that requires aggregating or filtering based on a simple predicate, because it avoids unnecessary data movement and extra passes.
OPTIMIZATION CHALLENGE
The key insight is recognizing that the predicate (value > K) is independent of other elements, allowing us to compute the answer on‑the‑fly without sorting, extra storage, or nested loops.
REAL-WORLD CONNECTION
Think of a network monitoring dashboard that continuously sums the load of servers exceeding a latency threshold; it processes each metric once, just like the algorithm processes each array element once.
During an interview, write the loop first, then immediately add the conditional check and accumulator; this demonstrates clarity of thought and avoids the temptation to over‑engineer with unnecessary data structures.
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. A naive solution might sort the array or use nested loops, which inflates time complexity to O(N log N) or O(N^2). The optimal paradigm leverages a single linear scan—often described as the two‑pointer or sliding‑window pattern—where we maintain a running total while iterating once through the list. This approach guarantees O(N) time because each element is examined exactly once, and O(1) auxiliary space since only a few scalar variables are needed.
Interview Questions on This Problem
Q1How would you modify the solution if you needed to return the count of elements greater than K instead of their sum?
Replace the accumulator that adds values with a simple counter that increments each time an element > K is encountered; the rest of the linear scan remains unchanged.
Q2Can you compute the sum of elements greater than K in a streaming fashion where the array size is unknown beforehand?
Yes—maintain a running sum and update it for each incoming element that exceeds K; this works because the operation is associative and requires no knowledge of future elements.
Q3If the input list is sorted in descending order, can you early‑terminate the scan? Explain the trade‑off.
In a descending list, once you encounter an element ≤ K you can break out because all subsequent elements will also be ≤ K, reducing average time. However, the worst‑case complexity remains O(N) and you must verify the sort order first, which may add overhead.
Examples
Input
5 1 4 7 2 9 5
Output
16
Explanation: The metrics greater than 5 are 7 and 9. Their sum is 7 + 9 = 16.
Input
3 -3 -1 -5 -2
Output
-1
Explanation: Only -1 is greater than -2. The sum is -1.
Input
6 10 20 30 40 50 60 25
Output
150
Explanation: Metrics greater than 25 are 30, 40, 50, and 60. Their sum is 30 + 40 + 50 + 60 = 150.
Constraints
- 1 <= N <= 100000
- -1000000000 <= metric[i] <= 1000000000
- -1000000000 <= K <= 1000000000
Optimal Approach & Strategy
Perform a single pass, checking each element against K and accumulating the sum when the condition holds.
Brute Force Approach
Sort the array then sum elements after finding the first element > K, or use nested loops to compare each element with every other.
Verified Code Solutions
/**
* @param {number[]} metrics
* @param {number} K
* @return {number}
*/
var detectNodes = function(metrics, K) {
let sum = 0;
for (let i = 0; i < metrics.length; i++) {
if (metrics[i] > K) {
sum += metrics[i];
}
}
return sum;
};class Solution {
public:
int detectNodes(vector<int>& metrics, int K) {
int sum = 0;
for (int i = 0; i < metrics.size(); ++i) {
if (metrics[i] > K) {
sum += metrics[i];
}
}
return sum;
}
};class Solution {
public int detectNodes(int[] metrics, int K) {
int sum = 0;
for (int i = 0; i < metrics.length; i++) {
if (metrics[i] > K) {
sum += metrics[i];
}
}
return sum;
}
}class Solution:
def detectNodes(self, metrics: List[int], K: int) -> int:
return sum(x for x in metrics if x > K)/**
* @param {number[]} metrics
* @param {number} K
* @return {number}
*/
var detectNodes = function(metrics, K) {
let sum = 0;
for (let i = 0; i < metrics.length; i++) {
if (metrics[i] > K) {
sum += metrics[i];
}
}
return sum;
};Asked in Top Tech Interviews
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