Network Node Detector 37 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing network and node metrics, construct an optimal algorithm to compute the target detector value under given operational constraints. Given an array of integers and an integer K, return the sum of all elements in the array that are less than or equal to K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Node Detector 37"
WHY DOES IT MATTER?
Filtering and aggregating data in one pass is a fundamental pattern for high‑throughput systems.
OPTIMIZATION CHALLENGE
The key is reducing the naive quadratic scan to a linear scan, cutting runtime by orders of magnitude.
REAL-WORLD CONNECTION
Network monitoring tools often sum metrics that fall below a threshold to detect anomalies.
Keep the loop tight, avoid extra data structures, and always use a type that safely holds the total.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The problem reduces to a single-pass filter: for each element we decide in O(1) whether it contributes to the sum based on the comparison with K. A naïve double‑loop that checks every pair of elements would be O(n²) and quickly exceeds time limits for large n, while the linear scan leverages the additive property of sums and the fact that each element is independent of others, making O(n) the optimal lower bound for an unsorted array. If the array were sorted, a binary search could locate the cutoff index and a prefix‑sum array would answer queries in O(log n), but for a single‑query scenario the overhead outweighs benefits. Thus the optimal paradigm is a straightforward linear traversal with constant extra space.
Interview Questions on This Problem
Q1What is the time and space complexity of summing elements ≤ K in an unsorted array?
Time complexity is O(n) because each element is inspected once; space complexity is O(1) as only a running total is stored.
Q2How would you modify the solution if you needed to answer many queries with different K values efficiently?
Sort the array and build a prefix‑sum array, then each query can be answered with a binary search in O(log n) time.
Q3Why might integer overflow be a concern, and how can you mitigate it in languages like Java or C++?
The sum may exceed the range of a 32‑bit int; use a larger type such as long long or BigInteger to store the accumulator.
Examples
Input
[1, 2, 3, 4, 5], 0
Output
0
Explanation: Step 1: Given the input array [1, 2, 3, 4, 5] and K = 0, we need to sum up values less than or equal to K. However, there are no values in the array that are less than or equal to 0. Therefore, the sum is 0.
Input
[10, 20, 30, 40, 50], 10
Output
60
Explanation: Step 1: Given the input array [10, 20, 30, 40, 50] and K = 10, we need to sum up values less than or equal to K. The values 10, 20, and 30 are less than or equal to K. Therefore, the sum is 10 + 20 + 30 = 60.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
The optimal approach uses a single loop with a conditional accumulator, achieving O(n) time and O(1) extra space.
Brute Force Approach
A brute‑force method would nest loops to compare every element with every other, resulting in O(n²) 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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