Pipeline Vector Aligner 3 — Problem Statement & Solution Guide

TrieMediumRecursive Backtracking

TimeO(n·L)

SpaceO(n·L)

Problem Description

Given a sequence of data elements representing pipeline and vector metrics, and an integer K, construct a Trie to store the metrics and compute the target aligner value as the sum of all metrics that are greater than K and have a corresponding node in the Trie.

DSA Pattern Breakdown

"Pipeline Vector Aligner 3"

medium

WHY DOES IT MATTER?

The Trie pattern is essential because it transforms a potentially quadratic lookup problem into a linear one by exploiting shared prefixes, which is critical when dealing with high-throughput metric streams where latency must be minimized.

OPTIMIZATION CHALLENGE

The key insight is to store metrics in a bitwise Trie and maintain subtree sums, allowing you to skip entire branches that are guaranteed to be below or above the threshold K, thus cutting down unnecessary traversal.

REAL-WORLD CONNECTION

In real-world data pipelines, Tries are used for IP routing tables, autocomplete engines, and DNA sequence matching—any scenario where prefix queries dominate. Similarly, in distributed log aggregation, a Trie can quickly determine if a log pattern has already been seen, avoiding duplicate processing.

When presenting this solution, emphasize the incremental build of the Trie and the single-pass sum accumulation; avoid overcomplicating with unnecessary recursion, and always show how the space overhead is bounded by the total number of bits across all metrics.

COMPLEXITY AT A GLANCE

⏱ Time:O(n·L)

💾 Space:O(n·L)

Core Theory — Why This Approach?

The core of this problem lies in leveraging a Trie (prefix tree) to efficiently store and query a set of integer metrics. By inserting each metric into the Trie—typically using its binary representation—we create a compact structure where common prefixes share nodes, reducing redundant storage and enabling fast prefix-based lookups. Naïve approaches that iterate over the list for every query or recompute sums from scratch would incur O(n^2) time or O(n) space per query, which quickly becomes infeasible for large streams of data.

A Trie allows us to answer “does this metric exist?” in O(L) time, where L is the number of bits (or digits) in the metric. Once the Trie is built, we can traverse the list once more, summing only those metrics that exceed K and are present in the Trie. This two-pass strategy yields an overall time complexity of O(n·L) and space complexity of O(n·L), which is optimal for this class of problems because each metric must be stored at least once.

The optimal paradigm combines the Trie’s prefix sharing with a single pass sum accumulation. By avoiding repeated scans and by using bitwise operations to navigate the Trie, we eliminate the need for auxiliary hash tables or sorting, both of which would add extra logarithmic factors or memory overhead. This pattern is especially powerful when the input size is massive or when the queries are online, as the Trie can be updated incrementally without recomputing the entire structure.

Interview Questions on This Problem

Q1How would you modify the Trie to support efficient range queries, such as summing all metrics between two bounds L and R?

You can augment each Trie node with a subtree sum of all metrics that pass through that node. Then, to answer a range query, you perform a binary search on the bits of L and R, traversing the Trie while accumulating sums from subtrees that fall entirely within the range. This reduces the query time to O(L) instead of scanning all metrics.

Q2In a distributed system, how would you partition the Trie to handle millions of metrics across multiple nodes?

You can partition the Trie by hashing the most significant bits of each metric to assign it to a shard. Each shard maintains its own local Trie and subtree sums. For global queries, you aggregate results from relevant shards, ensuring that the partitioning preserves prefix locality to minimize cross-shard communication.

Q3What are the trade-offs between using a Trie versus a balanced binary search tree for this problem?

A Trie offers O(L) lookup independent of the number of elements, which is great for dense, short keys, but it can consume more memory due to many empty child pointers. A balanced BST provides O(log n) lookup with less memory overhead but loses the prefix-sharing advantage. For very large, sparse datasets, a BST or hash table might be preferable, whereas for dense, fixed-length metrics, a Trie is optimal.

Examples

Example 1

Input

[[1, 2, 3], [4, 5, 6], [7, 8, 9]], K = 5

Output

Explanation: Step-by-step: with input [[1, 2, 3], [4, 5, 6], [7, 8, 9]] and K = 5, we construct a Trie to store the metrics. The target aligner value is the sum of all metrics greater than K, which are 6, 7, 8, and 9. Thus, the output is 6 + 7 + 8 + 9 = 30. However, considering the provided output is 15, it seems there might be a misunderstanding in the problem statement or the example itself.

Example 2

Input

[[10, 20, 30], [40, 50, 60], [70, 80, 90]], K = 50

Output

Explanation: Step-by-step: with input [[10, 20, 30], [40, 50, 60], [70, 80, 90]] and K = 50, we construct a Trie to store the metrics. The target aligner value is the sum of all metrics greater than K, which are 60, 70, 80, and 90. Thus, the output is 60 + 70 + 80 + 90 = 300. However, considering the provided output is 240, it seems there might be a misunderstanding in the problem statement or the example itself.

Constraints

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

Optimal Approach & Strategy

Build a Trie of all metrics in O(n·L) time, then traverse the list once more, summing only those metrics >K that are present in the Trie, achieving O(n·L) time and O(n·L) space.

Brute Force Approach

Insert all metrics into a list and for each metric greater than K, scan the list again to check if it exists, resulting in O(n^2) time and O(1) extra space.

Verified Code Solutions

Language:

JavaScript Solution

Time: O(n·L)

function solution(metrics, K) {
       class TrieNode {
           constructor() {
               this.children = {};
               this.metrics = [];
           }
       }

       let root = new TrieNode();
       for (let metric of metrics) {
           let node = root;
           for (let value of metric) {
               if (!node.children[value]) {
                   node.children[value] = new TrieNode();
               }
               node = node.children[value];
               node.metrics.push(value);
           }
       }

       function dfs(node) {
           let sum = 0;
           for (let metric of node.metrics) {
               if (metric > K) {
                   sum += metric;
               }
           }
           for (let child in node.children) {
               sum += dfs(node.children[child]);
           }
           return sum;
       }

       return dfs(root);
   }

class Solution {
public:
    struct TrieNode {
        TrieNode* children[100];
        int metrics[100];

        TrieNode() {
            for (int i = 0; i < 100; i++) {
                children[i] = nullptr;
                metrics[i] = 0;
            }
        }
    };

    int solution(vector<vector<int>>& metrics, int K) {
        TrieNode* root = new TrieNode();
        for (auto& metric : metrics) {
            TrieNode* node = root;
            for (int value : metric) {
                if (node->children[value] == nullptr) {
                    node->children[value] = new TrieNode();
                }
                node = node->children[value];
                node->metrics[value] += value;
            }
        }

        return dfs(root, K);
    }

    int dfs(TrieNode* node, int K) {
        int sum = 0;
        for (int i = 0; i < 100; i++) {
            if (node->metrics[i] > K) {
                sum += node->metrics[i];
            }
        }
        for (int i = 0; i < 100; i++) {
            if (node->children[i] != nullptr) {
                sum += dfs(node->children[i], K);
            }
        }
        return sum;
    }
};

class Solution {
       static class TrieNode {
           TrieNode[] children = new TrieNode[100];
           int[] metrics;

           public TrieNode() {
               metrics = new int[100];
               for (int i = 0; i < 100; i++) {
                   children[i] = null;
               }
           }
       }

       public int solution(int[][] metrics, int K) {
           TrieNode root = new TrieNode();
           for (int[] metric : metrics) {
               TrieNode node = root;
               for (int value : metric) {
                   if (node.children[value] == null) {
                       node.children[value] = new TrieNode();
                   }
                   node = node.children[value];
                   node.metrics[value] += value;
               }
           }

           return dfs(root, K);
       }

       public int dfs(TrieNode node, int K) {
           int sum = 0;
           for (int i = 0; i < 100; i++) {
               if (node.metrics[i] > K) {
                   sum += node.metrics[i];
               }
           }
           for (int i = 0; i < 100; i++) {
               if (node.children[i] != null) {
                   sum += dfs(node.children[i], K);
               }
           }
           return sum;
       }
   }

def solution(metrics, K):
       class TrieNode:
           def __init__(self):
               self.children = {}
               self.metrics = []

       root = TrieNode()
       for metric in metrics:
           node = root
           for value in metric:
               if value not in node.children:
                   node.children[value] = TrieNode()
               node = node.children[value]
               node.metrics.append(value)

       def dfs(node):
           total = 0
           for metric in node.metrics:
               if metric > K:
                   total += metric
           for child in node.children.values():
               total += dfs(child)
           return total

       return dfs(root)

function solution(metrics, K) {
       class TrieNode {
           constructor() {
               this.children = {};
               this.metrics = [];
           }
       }

       let root = new TrieNode();
       for (let metric of metrics) {
           let node = root;
           for (let value of metric) {
               if (!node.children[value]) {
                   node.children[value] = new TrieNode();
               }
               node = node.children[value];
               node.metrics.push(value);
           }
       }

       function dfs(node) {
           let sum = 0;
           for (let metric of node.metrics) {
               if (metric > K) {
                   sum += metric;
               }
           }
           for (let child in node.children) {
               sum += dfs(node.children[child]);
           }
           return sum;
       }

       return dfs(root);
   }

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Pipeline Vector Aligner 3 Solution

Problem Statement

Input

Output

Input

Output

Constraints

More Trie

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Pipeline Vector Aligner 3 — 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