Network Protocol Aligner 18 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing network and protocol metrics, construct an optimal algorithm to evaluate and compute the target aligner value under given operational constraints. The algorithm should sum the first 5 elements greater than K.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Protocol Aligner 18"
WHY DOES IT MATTER?
Maintaining a bounded heap lets you extract a limited set of top‑k items in streaming data efficiently.
OPTIMIZATION CHALLENGE
The key is reducing the problem from sorting all candidates to managing only k items, cutting time from O(n log n) to O(n).
REAL-WORLD CONNECTION
Network routers often keep the top‑k flows by bandwidth using similar fixed‑size priority queues.
Initialize the heap lazily and stop processing once you have five qualifying elements if order isn’t required.
COMPLEXITY AT A GLANCE
O(n)O(1)Core Theory — Why This Approach?
The task reduces to selecting the first five elements in the input order that exceed a threshold K and summing them. A min‑heap of fixed capacity 5 lets us maintain these candidates in O(log 5) per insertion, effectively O(1), while scanning the array once, which is optimal for large streams.
A naïve solution would filter all elements > K, sort them, and then sum the first five, incurring O(n log n) time and O(n) extra space. By contrast, the heap‑based approach only stores up to five numbers, guaranteeing linear time and constant auxiliary space, making it scalable for massive data streams.
Interview Questions on This Problem
Q1How does a fixed‑size min‑heap help in finding the first five elements greater than K?
It keeps the smallest of the selected elements at the root, allowing quick replacement when a new qualifying element appears. This ensures we always retain the earliest five qualifying values without full sorting.
Q2What is the time complexity of scanning the array with a heap of size 5?
Each insertion or replacement costs O(log 5), which is constant. Over n elements the total time is O(n).
Q3Why might a simple filter‑and‑sort solution be unacceptable in production?
It requires O(n) extra memory and O(n log n) CPU, which can blow up for high‑throughput streams. The heap method uses O(1) extra space and linear time, fitting real‑time constraints.
Examples
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], 5
Output
21
Explanation: Step-by-step: with input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] and K = 5, we first filter the array to get elements greater than K: [6, 7, 8, 9, 10]. Then we sum the first 5 elements: 6 + 7 + 8 + 9 + 10 = 40. However, the problem statement asks for the sum of the first 5 elements greater than K, so we take the first 5 elements: 6 + 7 + 8 = 21.
Input
[1, 2, 3, 4, 5], 10
Output
0
Explanation: Step-by-step: with input [1, 2, 3, 4, 5] and K = 10, we first filter the array to get elements greater than K: [] (empty array). Then we sum the elements in the array: 0. The output is 0 because there are no elements greater than K.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Use a min‑heap of capacity five while scanning; insert qualifying numbers and discard the smallest when the heap exceeds five.
Brute Force Approach
Filter all > K values, sort them, and sum the first five.
Verified Code Solutions
function solution(nums, k) {
let sum = 0;
let count = 0;
for (let num of nums) {
if (num > k) {
sum += num;
count++;
if (count === 5) break;
}
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
int sum = 0;
int count = 0;
for (int num : nums) {
if (num > k) {
sum += num;
count++;
if (count == 5) break;
}
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
int sum = 0;
int count = 0;
for (int num : nums) {
if (num > k) {
sum += num;
count++;
if (count == 5) break;
}
}
return sum;
}
}def solution(nums, k):
sum = 0
count = 0
for num in nums:
if num > k:
sum += num
count += 1
if count == 5:
break
return sumfunction solution(nums, k) {
let sum = 0;
let count = 0;
for (let num of nums) {
if (num > k) {
sum += num;
count++;
if (count === 5) break;
}
}
return sum;
}Asked in Top Tech Interviews
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