Network Network Detector 49 — Problem Statement & Solution Guide
Problem Description
Network Network Detector 49
You are given a list of integer measurements collected from a network monitoring system and an integer threshold K. Your task is to calculate the sum of all measurements that are strictly greater than K. The input consists of the number of measurements N, followed by N space‑separated integers, and finally the integer K. Output a single integer representing the required sum.
Design an algorithm that runs in linear time with respect to N. A frequency hash map (or dictionary) can be employed to count occurrences of each distinct measurement, allowing the sum to be accumulated efficiently without revisiting elements multiple times.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Network Detector 49"
WHY DOES IT MATTER?
Selecting and aggregating based on a condition is a fundamental data‑processing pattern.
OPTIMIZATION CHALLENGE
Avoid sorting or extra storage; achieve O(N) time with O(1) extra space.
REAL-WORLD CONNECTION
Network monitoring tools filter out-of‑range metrics before alerting operators.
Initialize the accumulator to zero and use a simple for‑each loop to keep the code cache‑friendly.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The problem reduces to a linear scan where each element is compared against a threshold K and, if larger, added to an accumulator. This is a classic example of a selection‑sum pattern that can be solved in O(N) time using a single pass.
A naive approach might attempt sorting or building auxiliary data structures, which adds unnecessary O(N log N) overhead and extra memory. The optimal paradigm leverages the fact that order is irrelevant; only a direct comparison and accumulation are needed, yielding optimal linear time and constant space.
Interview Questions on This Problem
Q1What is the time complexity of summing elements greater than a threshold in an unsorted array?
It can be done in O(N) by scanning once. Sorting would increase it to O(N log N), which is unnecessary.
Q2How would you handle integer overflow when summing large measurements?
Use a wider integer type like 64‑bit long long in C++ or Python's arbitrary‑precision int. Alternatively, check for overflow before each addition.
Q3Can this problem be solved in parallel, and what would be the trade‑off?
Yes, split the array into chunks, compute partial sums in parallel, then combine them. The overhead of thread management may outweigh benefits for small N.
Examples
Input
5 2 5 8 1 6 4
Output
19
Explanation: The measurements are [2,5,8,1,6] and K=4. Values greater than 4 are 5, 8 and 6. Their sum is 5+8+6 = 19.
Input
5 -3 0 4 10 -1 0
Output
14
Explanation: Measurements: [-3,0,4,10,-1], K=0. Elements exceeding 0 are 4 and 10. Sum = 4+10 = 14.
Input
3 100 200 300 250
Output
300
Explanation: Measurements: [100,200,300], K=250. Only 300 is larger than 250, so the sum is 300.
Constraints
- 1 <= N <= 10^5
- -10^9 <= measurement[i] <= 10^9
- -10^9 <= K <= 10^9
- The sum fits within a 64‑bit signed integer.
Optimal Approach & Strategy
Iterate once, compare each element to K, and add qualifying values to a sum variable.
Brute Force Approach
Sort the array then iterate from the first element greater than K, summing the rest.
Verified Code Solutions
const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
terminal: false
});
let lines = [];
let lineCount = 0;
rl.on('line', (line) => {
lines.push(line);
lineCount++;
if (lineCount === 3) {
rl.close();
solve();
}
});
function solve() {
const N = parseInt(lines[0]);
const measurements = lines[1].split(' ').map(Number);
const K = parseInt(lines[2]);
let sum = 0;
for (let i = 0; i < N; i++) {
if (measurements[i] > K) {
sum += measurements[i];
}
}
console.log(sum);
}#include <iostream>
#include <vector>
using namespace std;
int main() {
int N;
cin >> N;
vector<int> measurements(N);
for (int i = 0; i < N; ++i) {
cin >> measurements[i];
}
int K;
cin >> K;
int sum = 0;
for (int i = 0; i < N; ++i) {
if (measurements[i] > K) {
sum += measurements[i];
}
}
cout << sum << endl;
return 0;
}import java.util.*;
public class Main {
public static void main(String[] args) {
Scanner scanner = new Scanner(System.in);
int N = scanner.nextInt();
int[] measurements = new int[N];
for (int i = 0; i < N; i++) {
measurements[i] = scanner.nextInt();
}
int K = scanner.nextInt();
int sum = 0;
for (int i = 0; i < N; i++) {
if (measurements[i] > K) {
sum += measurements[i];
}
}
System.out.println(sum);
scanner.close();
}
}import sys
def main():
input = sys.stdin.read
data = input().split()
N = int(data[0])
measurements = list(map(int, data[1:N+1]))
K = int(data[N+1])
result = sum(x for x in measurements if x > K)
print(result)
if __name__ == "__main__":
main()const readline = require('readline');
const rl = readline.createInterface({
input: process.stdin,
terminal: false
});
let lines = [];
let lineCount = 0;
rl.on('line', (line) => {
lines.push(line);
lineCount++;
if (lineCount === 3) {
rl.close();
solve();
}
});
function solve() {
const N = parseInt(lines[0]);
const measurements = lines[1].split(' ').map(Number);
const K = parseInt(lines[2]);
let sum = 0;
for (let i = 0; i < N; i++) {
if (measurements[i] > K) {
sum += measurements[i];
}
}
console.log(sum);
}Asked in Top Tech Interviews
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