Network Node Synthesizer 32 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing network and node metrics, construct an optimal algorithm to evaluate and compute the target synthesizer value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Node Synthesizer 32"
WHY DOES IT MATTER?
Maintaining a bounded heap prevents unnecessary work on irrelevant data, crucial for real‑time analytics.
OPTIMIZATION CHALLENGE
The key is reducing the per‑element cost from O(log n) to O(log k) by limiting heap size.
REAL-WORLD CONNECTION
Network routers often keep the top‑k flows by bandwidth using a min‑heap to decide which flows to monitor.
Initialize the heap with the first k elements, then only push‑pop when a new metric exceeds the heap root.
COMPLEXITY AT A GLANCE
O(n log k)O(k)Core Theory — Why This Approach?
Heap data structures provide O(log n) insertion and extraction, making them ideal for maintaining a dynamic set of the most extreme values. By using a min‑heap of size k we can keep the k largest (or smallest) network metrics while scanning the input once, discarding any element that cannot affect the final answer.
A naive solution would sort the entire sequence, costing O(n log n) time and O(n) space, which quickly becomes prohibitive for massive telemetry streams. The optimal paradigm leverages the heap’s bounded size to achieve O(n log k) time and O(k) auxiliary space, guaranteeing scalability even when n reaches millions.
Interview Questions on This Problem
Q1Why is a min‑heap preferred over a max‑heap when extracting the k‑largest elements?
A min‑heap keeps the smallest of the top‑k at its root, allowing O(1) checks to discard smaller elements. This ensures the heap never exceeds size k, preserving the O(n log k) bound.
Q2How does the heap‑based solution degrade to O(n log n) if k equals n?
When k = n the heap grows to size n, so each insertion costs O(log n) and we perform n of them, matching full sort complexity. The space also rises to O(n), losing the advantage of a bounded heap.
Q3Can you retrieve the k‑largest values in sorted order directly from the heap?
Extracting repeatedly from a min‑heap yields values in ascending order, so you must either store them and reverse or use a max‑heap for descending order. Both approaches add an extra O(k log k) step after the main scan.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
49
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K=3, we first filter the array to include elements greater than or equal to K. This gives us [4, 5, 6, 7, 8, 9, 10]. We then calculate the sum of these elements, which is 4 + 5 + 6 + 7 + 8 + 9 + 10 = 49. Therefore, the output is 49.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
Output
40
Explanation: Step-by-step: Given the input array [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K=4, we first filter the array to include elements greater than or equal to K. This gives us [4, 5, 6, 7, 8, 9, 10]. We then calculate the sum of these elements, which is 4 + 5 + 6 + 7 + 8 + 9 + 10 = 49. However, we are only interested in the elements greater than or equal to K=4, so we exclude 4 and 5 from the sum. This gives us 6 + 7 + 8 + 9 + 10 = 40. Therefore, the output is 40.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a min‑heap of capacity k, insert the first k items, then for each remaining element replace the root if it is larger, achieving O(n log k) time and O(k) space.
Brute Force Approach
Sort the entire sequence and then pick the last k elements, which costs O(n log n) time and O(n) space.
Verified Code Solutions
function solution(nums, K) {
if (nums.length === 0 || nums.length === 1) return 0;
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) {
if (nums.size() == 0 || nums.size() == 1) return 0;
int sum = 0;
for (int num : nums) {
if (num >= K) sum += num;
}
return sum;
}
};class Solution {
public int solution(int[] nums, int K) {
if (nums.length == 0 || nums.length == 1) return 0;
int sum = 0;
for (int num : nums) {
if (num >= K) sum += num;
}
return sum;
}
}def solution(nums, K):
if len(nums) == 0 or len(nums) == 1:
return 0
sum = 0
for num in nums:
if num >= K:
sum += num
return sumfunction solution(nums, K) {
if (nums.length === 0 || nums.length === 1) return 0;
let sum = 0;
for (let num of nums) {
if (num >= K) sum += num;
}
return sum;
}Asked in Top Tech Interviews
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