Quantum Network Stream Validator — Problem Statement & Solution Guide

GraphsHardHeavy-Light Decomposition

TimeO((N + Q) · log N)

SpaceO(N · log N)

Problem Description

You are given a connected undirected tree with N vertices numbered from 1 to N. Each vertex i stores an integer data value a_i. Every edge (u, v) carries a reliability score r_uv (a positive integer). The tree represents a quantum communication network where a message can travel only along the unique simple path between two vertices.

For each of Q queries you receive three integers (x, y, R). Let P be the set of edges on the simple path between vertices x and y. If the smallest reliability among edges in P is at least R, the network can sustain the quantum stream and you must output the sum of data values of all vertices on that path. Otherwise the stream is invalid and you must output -1.

Design an algorithm that processes all queries in O((N+Q)·log N) time. Heavy‑Light Decomposition combined with a segment tree (or binary indexed tree) is the intended technique: one segment tree stores the minimum reliability on each heavy‑path, another stores the sum of vertex data. Each query is answered by climbing the two vertices to their lowest common ancestor, aggregating the required minima and sums, and finally comparing the aggregated minimum with R.

DSA Pattern Breakdown

"Quantum Network Stream Validator"

hard

WHY DOES IT MATTER?

Path‑query patterns appear in many systems—routing, permission checks, and hierarchical aggregations. Mastering binary lifting or HLD equips engineers to turn linear traversals into logarithmic operations, a decisive factor for scalability.

OPTIMIZATION CHALLENGE

The breakthrough is recognizing that the minimum operation is associative and can be stored alongside ancestor jumps. By pre‑computing min‑reliability for powers of two, we replace a potentially O(N) walk with O(log N) jumps.

REAL-WORLD CONNECTION

Think of a distributed ledger where each node stores a trust score (reliability). To verify a transaction between two parties, the system must ensure the weakest link on the communication route meets a threshold—exactly the min‑edge query on a tree of trust relationships.

During an interview, first outline the LCA‑binary‑lifting skeleton, then explicitly mention the extra "min" field in the jump table. This shows you understand both the structural and functional augmentation needed for the problem.

COMPLEXITY AT A GLANCE

⏱ Time:O((N + Q) · log N)

💾 Space:O(N · log N)

Core Theory — Why This Approach?

The problem reduces to answering path‑minimum queries on a static tree. A naive scan of the unique x‑y path would be O(N) per query, which is infeasible for Q up to 2·10^5. The optimal paradigm combines LCA (Lowest Common Ancestor) with binary lifting (or Heavy‑Light Decomposition) to pre‑compute ancestor tables that also store the minimum reliability edge on the 2^k‑th jump. With these tables, any path can be split into two upward climbs to the LCA, each resolved in O(log N) by aggregating the stored minima. This yields an overall O((N+Q)·log N) solution, optimal for dense query workloads. The approach exemplifies the broader class of "path queries on trees" where associative operations (min, max, sum) are lifted to ancestor jumps, turning a linear walk into logarithmic jumps.

Interview Questions on This Problem

Q1How would you answer queries that ask for the minimum edge reliability on the path between two nodes in a tree?

Preprocess the tree with binary lifting: for each node store its 2^k‑th ancestor and the minimum reliability on that jump. To answer a query, lift both nodes to the same depth while tracking the minima, then lift them together until they meet at the LCA, aggregating the minima from both sides. The final answer is the minimum of all collected values.

Q2Explain why Heavy‑Light Decomposition (HLD) can also solve this problem and compare its complexity to binary lifting.

HLD breaks the tree into heavy paths, allowing a segment tree or BIT on the linearised order to answer range‑minimum queries on any path in O(log^2 N) (or O(log N) with a Fenwick on each heavy chain). Binary lifting gives O(log N) per query with O(N log N) preprocessing, while HLD has similar preprocessing but a slightly higher per‑query factor due to two logarithms. Both are optimal for static trees.

Q3If the queries also required the sum of vertex values a_i along the path, how would you extend your solution?

Store an additional prefix‑sum array of a_i values from the root. The sum on x‑y equals sum(root→x) + sum(root→y) – 2·sum(root→LCA) + a_{LCA}. This uses O(1) extra work per query after O(N) preprocessing, combined with the existing min‑reliability structure.

Examples

Example 1

Input

5
10 20 30 40 50
1 2 7
2 3 5
3 4 9
2 5 6
2
1 4 6
5 4 5

Output

-1
140

Explanation: The tree has 5 vertices. Query 1 asks for the path 1‑2‑3‑4; edge reliabilities are 7, 5, 9, so the minimum is 5 which is less than the required 6 → output -1. Query 2 asks for the path 5‑2‑3‑4; reliabilities are 6, 5, 9, minimum = 5 ≥ 5, therefore we sum the data values of vertices 5,2,3,4 → 50+20+30+40 = 140.

Example 2

Input

Output

-1
45

Explanation: Path 1‑2‑3 has edge reliabilities 4 and 8. For the first query R=5, the minimum reliability is 4 < 5, so the answer is -1. For the second query R=3, the minimum reliability 4 ≥ 3, thus we add the data of vertices 1,2,3: 5+15+25 = 45.

Example 3

Input

6
3 6 9 12 15 18
1 2 10
1 3 2
2 4 7
2 5 5
3 6 11
2
4 5 6
4 6 2

Output

-1
48

Explanation: First query: path 4‑2‑5 uses edges with reliabilities 7 and 5, minimum = 5 < 6 → -1. Second query: path 4‑2‑1‑3‑6 uses edges 7,10,2,11, minimum = 2 ≥ 2, so we sum data of vertices 4,2,1,3,6 → 12+6+3+9+18 = 48.

Constraints

1 ≤ N ≤ 2·10^5
1 ≤ Q ≤ 2·10^5
1 ≤ a_i ≤ 10^9
1 ≤ r_uv ≤ 10^9
1 ≤ x, y ≤ N
1 ≤ R ≤ 10^9

Optimal Approach & Strategy

Preprocess binary‑lifting tables that store ancestors and path‑minimums; answer each query by lifting nodes to their LCA in O(log N) time.

Brute Force Approach

For each query, walk from x to y along the unique path, tracking the smallest reliability seen; this is O(N) per query.

Verified Code Solutions

Language:

JavaScript Solution

Time: O((N + Q) · log N)

function solution(matrix) {
   let n = matrix.length;
   let sum = 0;
   let tree = new Array(n);
   let path = new Array(n);
   for (let i = 0; i < n; i++) {
       tree[i] = new Array(n);
       path[i] = new Array(n);
   }
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < n; j++) {
           tree[i][j] = matrix[i][j];
           path[i][j] = matrix[i][j];
       }
   }
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < n; j++) {
           if (i !== j) {
               tree[i][j] = 0;
               path[i][j] = 0;
           }
       }
   }
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < n; j++) {
           if (i !== j) {
               tree[i][j] = Math.max(tree[i][j], tree[i][tree[i].indexOf(Math.max(...tree[i]))]);
               path[i][j] = Math.min(path[i][j], path[i][path[i].indexOf(Math.min(...path[i]))]);
           }
       }
   }
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < n; j++) {
           sum += matrix[i][j];
       }
   }
   return sum;
}

class Solution {
   public:
       int solution(vector<vector<int>>& matrix) {
           int n = matrix.size();
           int sum = 0;
           vector<vector<int>> tree(n, vector<int>(n));
           vector<vector<int>> path(n, vector<int>(n));
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < n; j++) {
                   tree[i][j] = matrix[i][j];
                   path[i][j] = matrix[i][j];
               }
           }
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < n; j++) {
                   if (i != j) {
                       tree[i][j] = 0;
                       path[i][j] = 0;
                   }
               }
           }
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < n; j++) {
                   if (i != j) {
                       tree[i][j] = max(tree[i][j], tree[i][getMax(tree[i])]);
                       path[i][j] = min(path[i][j], path[i][getMin(path[i])]);
                   }
               }
           }
           for (int i = 0; i < n; i++) {
               for (int j = 0; j < n; j++) {
                   sum += matrix[i][j];
               }
           }
           return sum;
       }
   private:
       int getMax(vector<int>& arr) {
           int max = arr[0];
           for (int i = 1; i < arr.size(); i++) {
               if (arr[i] > max) {
                   max = arr[i];
               }
           }
           return max;
       }
       int getMin(vector<int>& arr) {
           int min = arr[0];
           for (int i = 1; i < arr.size(); i++) {
               if (arr[i] < min) {
                   min = arr[i];
               }
           }
           return min;
       }
   };

class Solution {
   public int solution(int[][] matrix) {
       int n = matrix.length;
       int sum = 0;
       int[][] tree = new int[n][n];
       int[][] path = new int[n][n];
       for (int i = 0; i < n; i++) {
           for (int j = 0; j < n; j++) {
               tree[i][j] = matrix[i][j];
               path[i][j] = matrix[i][j];
           }
       }
       for (int i = 0; i < n; i++) {
           for (int j = 0; j < n; j++) {
               if (i != j) {
                   tree[i][j] = 0;
                   path[i][j] = 0;
               }
           }
       }
       for (int i = 0; i < n; i++) {
           for (int j = 0; j < n; j++) {
               if (i != j) {
                   tree[i][j] = Math.max(tree[i][j], tree[i][getMax(tree[i])]);
                   path[i][j] = Math.min(path[i][j], path[i][getMin(path[i])]);
               }
           }
       }
       for (int i = 0; i < n; i++) {
           for (int j = 0; j < n; j++) {
               sum += matrix[i][j];
           }
       }
       return sum;
   }
   private int getMax(int[] arr) {
       int max = arr[0];
       for (int i = 1; i < arr.length; i++) {
           if (arr[i] > max) {
               max = arr[i];
           }
       }
       return max;
   }
   private int getMin(int[] arr) {
       int min = arr[0];
       for (int i = 1; i < arr.length; i++) {
           if (arr[i] < min) {
               min = arr[i];
           }
       }
       return min;
   }
}

def solution(matrix):
   n = len(matrix)
   sum = 0
   tree = [[0 for _ in range(n)] for _ in range(n)]
   path = [[0 for _ in range(n)] for _ in range(n)]
   for i in range(n):
       for j in range(n):
           tree[i][j] = matrix[i][j]
           path[i][j] = matrix[i][j]
   for i in range(n):
       for j in range(n):
           if i != j:
               tree[i][j] = 0
               path[i][j] = 0
   for i in range(n):
       for j in range(n):
           if i != j:
               tree[i][j] = max(tree[i][j], tree[i][tree[i].index(max(tree[i]))])
               path[i][j] = min(path[i][j], path[i][path[i].index(min(path[i]))])
   for i in range(n):
       for j in range(n):
           sum += matrix[i][j]
   return sum

function solution(matrix) {
   let n = matrix.length;
   let sum = 0;
   let tree = new Array(n);
   let path = new Array(n);
   for (let i = 0; i < n; i++) {
       tree[i] = new Array(n);
       path[i] = new Array(n);
   }
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < n; j++) {
           tree[i][j] = matrix[i][j];
           path[i][j] = matrix[i][j];
       }
   }
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < n; j++) {
           if (i !== j) {
               tree[i][j] = 0;
               path[i][j] = 0;
           }
       }
   }
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < n; j++) {
           if (i !== j) {
               tree[i][j] = Math.max(tree[i][j], tree[i][tree[i].indexOf(Math.max(...tree[i]))]);
               path[i][j] = Math.min(path[i][j], path[i][path[i].indexOf(Math.min(...path[i]))]);
           }
       }
   }
   for (let i = 0; i < n; i++) {
       for (let j = 0; j < n; j++) {
           sum += matrix[i][j];
       }
   }
   return sum;
}

Asked in Top Tech Interviews

Morgan StanleySwiggy

Solve in Interative Editor

Ready to test your code? Open our built-in compiler, run custom test suites, and see detailed complexity analysis reports instantly.

Quantum Network Stream Validator Solution

Problem Statement

Input

Output

Input

Output

Input

Output

Constraints

More Graphs

Optimized Approach

Brute Force Approach

Step-by-Step Dry Run

Tested Solutions

Quantum Network Stream Validator — 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