Network Node Extractor 16 — Problem Statement & Solution Guide
Problem Description
You are given a list of integer metrics that describe the performance of nodes in a communication network. A threshold value K is also provided. Your task is to compute the total of all metrics that are greater than or equal to K.
Input
The first line contains an integer N, the number of metrics. The second line contains N space‑separated integers representing the metrics. The third line contains the integer K.
Output
Print a single integer: the sum of all metrics that satisfy the condition metric ≥ K.
The algorithm should run in linear time with respect to N and use only constant additional memory.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Node Extractor 16"
WHY DOES IT MATTER?
Linear‑time filtering is a fundamental pattern for processing streams of data efficiently.
OPTIMIZATION CHALLENGE
The key is to avoid sorting or extra passes, reducing complexity from O(N log N) to O(N).
REAL-WORLD CONNECTION
Network monitoring tools constantly sum metrics that exceed alert thresholds in real time.
Implement a single‑pass loop that updates the sum directly, and reserve a stack only if you need to preserve qualifying values for later use.
COMPLEXITY AT A GLANCE
O(N)O(1)Core Theory — Why This Approach?
The task reduces to a classic filtering‑and‑aggregation problem: we must examine each metric and add it to a running total if it meets the threshold K. A naïve approach might sort the array and then binary‑search for the first qualifying element, which incurs O(N log N) time and unnecessary overhead for a problem that only requires a linear scan. The optimal paradigm leverages the fact that any solution must at least look at each element once, yielding an O(N) time algorithm that is provably optimal for unsorted input.
While the problem is categorized under the Stack topic, a stack can be employed to temporarily hold qualifying metrics before summation, but this adds O(N) auxiliary space without benefit. The most efficient implementation simply accumulates the sum on the fly, using constant extra space, which aligns with streaming‑style processing common in real‑time systems.
Interview Questions on This Problem
Q1What is the time complexity of summing elements that satisfy a condition in an unsorted array?
It is O(N) because each element must be inspected at least once. Any algorithm faster would miss potential qualifying elements.
Q2Why might sorting the array before summing be a suboptimal choice here?
Sorting adds O(N log N) overhead, which is unnecessary when a single pass suffices. The extra work does not reduce the overall work of checking each element.
Q3How can you handle potential integer overflow when summing large metrics?
Use a wider integer type such as 64‑bit long long in C++ or Python's arbitrary‑precision int. Accumulate in that type to avoid overflow.
Examples
Input
5 1 2 3 4 5 3
Output
12
Explanation: Metrics that are at least 3 are 3, 4, and 5. Their sum is 3+4+5=12.
Input
4 -5 -2 0 3 0
Output
3
Explanation: Metrics ≥0 are 0 and 3. Sum is 0+3=3.
Input
6 10 20 30 40 50 60 35
Output
150
Explanation: Metrics ≥35 are 40, 50, and 60. Sum is 40+50+60=150.
Constraints
- 1 <= N <= 100000
- -1000000000 <= nums[i] <= 1000000000
- -1000000000 <= K <= 1000000000
Optimal Approach & Strategy
The optimal solution scans the array once, conditionally adds to a sum, achieving O(N) time and O(1) space.
Brute Force Approach
A brute‑force method might sort the array then sum from the first element >= K, costing O(N log N) time.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (num >= k) {
sum += num;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sum = 0;
for (int num : nums) {
if (num >= k) {
sum += num;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
for (int num : nums) {
if (num >= k) {
sum += num;
}
}
return sum;
}
}def solution(nums, k):
sum = 0
for num in nums:
if num >= k:
sum += num
return sumfunction solution(nums, k) {
let sum = 0;
for (let num of nums) {
if (num >= k) {
sum += num;
}
}
return sum;
}Asked in Top Tech Interviews
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