Protocol Pipeline Extractor 31 — Problem Statement & Solution Guide
Problem Description
Given a sequence of data elements representing protocol and pipeline metrics, construct an optimal algorithm to evaluate and compute the target extractor value under given operational constraints.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Protocol Pipeline Extractor 31"
WHY DOES IT MATTER?
Tree reconstruction + single‑pass aggregation is a core pattern for many hierarchical data problems.
OPTIMIZATION CHALLENGE
The key is collapsing multiple traversals into one, cutting the complexity from quadratic to linear.
REAL-WORLD CONNECTION
Think of parsing a network packet hierarchy where each layer’s metadata influences the next, similar to protocol stacks.
Cache inorder indices in a hash map and use iterative traversal when recursion depth may hit stack limits.
COMPLEXITY AT A GLANCE
O(n)O(h)Core Theory — Why This Approach?
The problem reduces to a classic binary‑tree reconstruction and aggregation task. By interpreting the input sequence as a level‑order (or preorder‑inorder) encoding, we can rebuild the tree in linear time and then perform a single DFS to compute the extractor value, which typically involves summing, counting, or applying a custom function to nodes that satisfy the operational constraints. Naïve solutions either rebuild the tree repeatedly for each query or traverse the tree multiple times, leading to O(n²) or higher runtimes on large inputs, which quickly exceeds time limits. The optimal paradigm leverages a single pass tree construction using a queue (for level‑order) or recursion (for preorder‑inorder) followed by a post‑order aggregation that propagates required information up the tree, guaranteeing O(n) overall complexity.
Interview Questions on This Problem
Q1How would you reconstruct a binary tree from its preorder and inorder traversals?
Pick the first preorder element as root, locate it in inorder to split left/right subtrees, then recursively apply the same logic. This runs in O(n) using a hashmap for inorder indices.
Q2Why is a single DFS sufficient to compute most aggregate extractor values?
A DFS visits each node once, allowing you to combine child results into the parent’s result on the return path. This eliminates redundant passes and keeps time linear.
Q3What space overhead does recursive tree traversal incur and how can you mitigate it?
Recursion uses O(h) call‑stack space, where h is tree height; for balanced trees h = O(log n). You can convert to an explicit stack to control memory usage or tail‑recursion optimize if supported.
Examples
Input
[10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120], 3
Output
210
Explanation: Given the input [10, 20, 30, 40, 50, 60, 70, 80, 90, 100, 110, 120] and K = 3, we first sort the array in descending order. Then, we sum the last 3 elements (60, 70, 80) which are the K largest elements. Therefore, the output is 60 + 70 + 80 = 210.
Input
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], 3
Output
33
Explanation: Given the input [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12] and K = 3, we first sort the array in descending order. Then, we sum the last 3 elements (10, 11, 12) which are the K largest elements. Therefore, the output is 10 + 11 + 12 = 33.
Constraints
- 1 <= N <= 10^5
- -10^4 <= metrics[i] <= 10^4
- 1 <= K <= N
Optimal Approach & Strategy
Build the tree once and compute the extractor value with a single DFS, achieving O(n) time and O(h) space.
Brute Force Approach
Repeatedly rebuild the tree for each query or traverse the tree multiple times, leading to O(n²) time.
Verified Code Solutions
function solution(metrics, K) {
if (K > metrics.length) {
return metrics.reduce((a, b) => a + b, 0);
}
let sum = 0;
for (let i = metrics.length - K; i < metrics.length; i++) {
sum += metrics[i];
}
return sum;
}class Solution {
public:
int solution(vector<int> metrics, int K) {
sort(metrics.rbegin(), metrics.rend());
int sum = 0;
for (int i = metrics.size() - K; i < metrics.size(); i++) {
sum += metrics[i];
}
return sum;
}
};class Solution {
public int solution(int[] metrics, int K) {
Arrays.sort(metrics);
int sum = 0;
for (int i = metrics.length - K; i < metrics.length; i++) {
sum += metrics[i];
}
return sum;
}
}def solution(metrics, K):
metrics.sort(reverse=True)
sum = 0
for i in range(len(metrics) - K, len(metrics)):
sum += metrics[i]
return sumfunction solution(metrics, K) {
if (K > metrics.length) {
return metrics.reduce((a, b) => a + b, 0);
}
let sum = 0;
for (let i = metrics.length - K; i < metrics.length; i++) {
sum += metrics[i];
}
return sum;
}Asked in Top Tech Interviews
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