Protocol Pipeline Aligner 45 — Problem Statement & Solution Guide

TrieEasyBFS / Union Find

TimeO(N·L + K log K)

SpaceO(N·L + K)

Problem Description

Given a sequence of data elements representing protocol and pipeline metrics, construct an optimal algorithm to evaluate and compute the target aligner value under given operational constraints. The output should be calculated based on the given K value, where K is the number of top metrics to consider.

DSA Pattern Breakdown

"Protocol Pipeline Aligner 45"

easy

WHY DOES IT MATTER?

Efficient top‑K extraction on hierarchical data prevents performance bottlenecks in real‑time monitoring systems.

OPTIMIZATION CHALLENGE

The key is reducing repeated O(N log N) sorts to a single linear‑time build plus a bounded heap operation.

REAL-WORLD CONNECTION

Network devices aggregate packet counters by protocol prefix, similar to how a Trie groups metric identifiers.

Cache the Trie across queries and only rebuild when the underlying metric stream changes.

COMPLEXITY AT A GLANCE

⏱ Time:O(N·L + K log K)

💾 Space:O(N·L + K)

Core Theory — Why This Approach?

A Trie (prefix tree) is ideal for aggregating hierarchical protocol identifiers because each node represents a shared prefix, allowing O(L) insertion and lookup where L is the identifier length. By storing a running count or metric sum at every node, we can answer “top‑K” queries across the entire dataset without re‑sorting the whole list, which would be O(N log N) for each query. Naïve approaches such as sorting the full array of metrics for every K request or scanning the list repeatedly lead to quadratic or near‑quadratic time on large inputs, quickly exhausting CPU and memory limits. The optimal paradigm builds the Trie once (O(N·L) time, O(N·L) space) and then performs a depth‑first traversal with a min‑heap of size K to extract the K highest aggregated values, guaranteeing O(N·L + K log K) overall performance.

Interview Questions on This Problem

Q1Why is a Trie preferred over sorting when repeatedly querying top‑K protocol metrics?

A Trie aggregates shared prefixes in O(L) per insertion, eliminating the need to re‑sort the entire dataset for each query. This reduces repeated O(N log N) work to a single O(N·L) build plus a cheap K‑heap extraction.

Q2How does maintaining a min‑heap of size K during a Trie traversal help achieve optimal time complexity?

The min‑heap keeps only the current K best values, so each node insertion or replacement costs O(log K). Traversing all nodes once yields O(N·L + K log K) instead of sorting all N values.

Q3What edge case must you handle when K exceeds the number of distinct metric groups in the Trie?

If K is larger than the available groups, the algorithm should return all aggregated values without error. Guarding against out‑of‑range heap operations prevents runtime exceptions.

Examples

Example 1

Input

["protocol1", "pipeline1", "protocol2", "pipeline2"] with K = 2

Output

The target aligner value based on the top 2 metrics

Explanation: Step-by-step: with input ["protocol1", "pipeline1", "protocol2", "pipeline2"] and K = 2, we first construct a Trie with the given protocol and pipeline metrics. Then, we evaluate the top 2 metrics based on their frequency or other given operational constraints, and finally compute the target aligner value.

Example 2

Input

["protocol1", "pipeline1", "protocol2", "pipeline2"] with K = 3

Output

The target aligner value based on the top 3 metrics, which in this case is all metrics since K is greater than or equal to the number of metrics

Explanation: Step-by-step: with input ["protocol1", "pipeline1", "protocol2", "pipeline2"] and K = 3, we follow the same process as before but consider all metrics since K equals the total number of metrics.

Constraints

1 <= N <= 10^5
-10^4 <= metrics[i] <= 10^4
1 <= K <= N

Optimal Approach & Strategy

Construct a Trie once, aggregate sums at nodes, then extract top K using a bounded min‑heap during a single traversal.

Brute Force Approach

Sort the entire list of metric values for every query and pick the first K entries, leading to O(N log N) per query.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(N·L + K log K)

function solution(metrics, K) {
       if (K > metrics.length) {
           K = metrics.length;
       }
       // Construct Trie and calculate target aligner value
       let trie = {};
       for (let metric of metrics) {
           let node = trie;
           for (let char of metric) {
               if (!node[char]) {
                   node[char] = {};
               }
               node = node[char];
           }
       }
       // Evaluate top K metrics
       let topMetrics = Object.keys(trie).sort((a, b) => {
           // Custom sorting based on operational constraints
           return b.localeCompare(a);
       }).slice(0, K);
       // Compute target aligner value
       let targetValue = 0;
       for (let metric of topMetrics) {
           targetValue += metric.length; // Example operation, replace with actual logic
       }
       return targetValue;
   }

class Solution {
public:
    int solution(vector<string> metrics, int K) {
        if (K > metrics.size()) {
            K = metrics.size();
        }
        // Construct Trie and calculate target aligner value
        unordered_map<char, unordered_map<string, int>> trie;
        for (const string& metric : metrics) {
            unordered_map<string, int>* node = &trie;
            for (char c : metric) {
                if (node->find(c) == node->end()) {
                    (*node)[c] = {};
                }
                node = &(*node)[c];
            }
        }
        // Evaluate top K metrics
        vector<string> topMetrics;
        for (const auto& pair : trie) {
            topMetrics.push_back(pair.first);
        }
        sort(topMetrics.begin(), topMetrics.end(), greater<string>());
        topMetrics.resize(K);
        // Compute target aligner value
        int targetValue = 0;
        for (const string& metric : topMetrics) {
            targetValue += metric.size(); // Example operation, replace with actual logic
        }
        return targetValue;
    }
};

import java.util.*;
class Solution {
    public int solution(String[] metrics, int K) {
        if (K > metrics.length) {
            K = metrics.length;
        }
        // Construct Trie and calculate target aligner value
        Map<Character, Map<String, Integer>> trie = new HashMap<>();
        for (String metric : metrics) {
            Map<String, Integer> node = trie;
            for (char c : metric.toCharArray()) {
                if (!node.containsKey(c)) {
                    node.put(c, new HashMap<>());
                }
                node = node.get(c);
            }
        }
        // Evaluate top K metrics
        List<String> topMetrics = new ArrayList<>(trie.keySet());
        Collections.sort(topMetrics, (a, b) -> b.compareTo(a));
        topMetrics = topMetrics.subList(0, K);
        // Compute target aligner value
        int targetValue = 0;
        for (String metric : topMetrics) {
            targetValue += metric.length(); // Example operation, replace with actual logic
        }
        return targetValue;
    }
}

def solution(metrics, K):
       if K > len(metrics):
           K = len(metrics)
       # Construct Trie and calculate target aligner value
       trie = {}
       for metric in metrics:
           node = trie
           for char in metric:
               if char not in node:
                   node[char] = {}
               node = node[char]
       # Evaluate top K metrics
       top_metrics = sorted(trie.keys(), reverse=True)[:K]
       # Compute target aligner value
       target_value = 0
       for metric in top_metrics:
           target_value += len(metric)  # Example operation, replace with actual logic
       return target_value

function solution(metrics, K) {
       if (K > metrics.length) {
           K = metrics.length;
       }
       // Construct Trie and calculate target aligner value
       let trie = {};
       for (let metric of metrics) {
           let node = trie;
           for (let char of metric) {
               if (!node[char]) {
                   node[char] = {};
               }
               node = node[char];
           }
       }
       // Evaluate top K metrics
       let topMetrics = Object.keys(trie).sort((a, b) => {
           // Custom sorting based on operational constraints
           return b.localeCompare(a);
       }).slice(0, K);
       // Compute target aligner value
       let targetValue = 0;
       for (let metric of topMetrics) {
           targetValue += metric.length; // Example operation, replace with actual logic
       }
       return targetValue;
   }

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Protocol Pipeline Aligner 45 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Trie

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Protocol Pipeline Aligner 45 — Problem Statement & Solution Guide

Problem Description

DSA Pattern Breakdown

Core Theory — Why This Approach?

Interview Questions on This Problem

Examples

Constraints

Optimal Approach & Strategy

Brute Force Approach

Verified Code Solutions

Asked in Top Tech Interviews

Solve in Interative Editor