Network Node Evaluator 40 — 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 evaluator value under given operational constraints. The operational constraints are that the function should add the K largest values.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Network Node Evaluator 40"
WHY DOES IT MATTER?
Selecting K largest items appears in ranking, recommendation, and resource‑allocation systems.
OPTIMIZATION CHALLENGE
The challenge is to avoid O(N log N) sorting by exploiting structural ordering of the data.
REAL-WORLD CONNECTION
Search engines use prefix trees to quickly surface top‑ranked URLs matching a query prefix.
Build the Trie once, then reuse it for multiple K‑queries; cache the traversal state to cut repeated work.
COMPLEXITY AT A GLANCE
O(N·L)O(N·L)Core Theory — Why This Approach?
A Trie (prefix tree) stores sequences (e.g., binary or decimal representations of numbers) in a hierarchical manner, allowing O(L) insertion and query where L is the length of the representation. By traversing the Trie from the most significant digit to the least, we can enumerate numbers in descending order without sorting the entire dataset, which is crucial when we only need the K largest values.
Naïve solutions either sort the whole array (O(N log N)) or scan repeatedly to find the maximum (O(K·N)), both of which become prohibitive for large N. The optimal paradigm combines a linear‑time Trie build (O(N·L)) with a controlled depth‑first search that stops after collecting K elements, yielding O(N·L + K·L) ≈ O(N·L) time and drastically reducing unnecessary work.
Interview Questions on This Problem
Q1How does a Trie enable retrieval of the K largest numeric values without full sorting?
Because numbers are stored digit‑by‑digit, a depth‑first walk that prefers larger digits visits values in descending order, allowing early termination after K hits.
Q2What is the time complexity of building a Trie for N integers with at most D digits?
Insertion is O(N·D) since each integer contributes D nodes at most.
Q3When would a min‑heap be preferable to a Trie for the K‑largest problem?
If the input is streaming or memory‑constrained, a min‑heap maintains only K elements in O(N log K) time without storing all digits.
Examples
Input
[100, 90, 80, 70, 60, 50, 40, 30, 20, 10, 5]
Output
270
Explanation: Step-by-step: Given the input [100, 90, 80, 70, 60, 50, 40, 30, 20, 10, 5] and K = 3, we first sort the array in descending order. Then, we sum the first three elements: 100 + 90 + 80 = 270.
Input
[5, 4, 3, 2, 1]
Output
5
Explanation: Step-by-step: Given the input [5, 4, 3, 2, 1] and K = 1, we first sort the array in descending order. Then, we sum the first element: 5.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Insert all numbers into a digit Trie and perform a descending DFS that stops after K values, achieving O(N·L) time.
Brute Force Approach
Sort the entire array and sum the last K elements, which costs O(N log N) time.
Verified Code Solutions
function solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k && i < nums.length; i++) {
sum += nums[i];
}
return sum;
}class Solution {
public:
int solution(vector<int>& nums, int k) {
sort(nums.begin(), nums.end(), greater<int>());
int sum = 0;
for (int i = 0; i < k && i < nums.size(); i++) {
sum += nums[i];
}
return sum;
}
};class Solution {
public int solution(int[] nums, int k) {
Arrays.sort(nums);
int sum = 0;
for (int i = 0; i < k && i < nums.length; i++) {
sum += nums[i];
}
return sum;
}
}def solution(nums, k):
nums.sort(reverse=True)
sum = 0
for i in range(k):
if i < len(nums):
sum += nums[i]
return sumfunction solution(nums, k) {
nums.sort((a, b) => b - a);
let sum = 0;
for (let i = 0; i < k && i < nums.length; i++) {
sum += nums[i];
}
return sum;
}Asked in Top Tech Interviews
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