Quantum Network Stream Evaluator 5 — Problem Statement & Solution Guide

GraphsHardHeavy-Light Decomposition

TimeO((N+Q) log N)

SpaceO(N)

Problem Description

You are given a tree with N vertices numbered from 1 to N. Each vertex i holds an integer value a_i. The tree is described by N‑1 undirected edges, each connecting two distinct vertices. After building the tree you must process Q queries of two possible forms:

1 u x – assign the value of vertex u to x.

2 u v – compute the sum of the values of all vertices that lie on the unique simple path between u and v (including u and v themselves) and output this sum.

All updates and path‑sum queries must be answered online. An efficient solution should run in O((N+Q)·log N) time, for example by applying Heavy‑Light Decomposition together with a segment tree or binary‑indexed tree on the linearised chains.

DSA Pattern Breakdown

"Quantum Network Stream Evaluator 5"

hard

WHY DOES IT MATTER?

Tree path queries are a classic example of a non‑linear data structure problem where naive traversal is too slow. Decomposing the tree into linear segments allows the use of efficient range query structures, turning a geometric problem into a well‑understood array problem.

OPTIMIZATION CHALLENGE

The key insight is that any simple path in a tree can be represented as a union of O(log N) contiguous segments after a suitable decomposition. This reduces the problem from potentially O(N) per query to O(log N) by leveraging segment trees or Fenwick trees.

REAL-WORLD CONNECTION

Consider a network of routers where each router holds a traffic load. Updating a router’s load and querying the total traffic along a route between two routers is analogous to point updates and path sums on a tree. Efficiently answering such queries is critical for real‑time network monitoring and load balancing.

When explaining the solution in an interview, emphasize the two‑step process: (1) map tree nodes to an array via HLD/Euler Tour, (2) use a segment tree for point updates and range queries. Highlight that the heavy path decomposition guarantees at most log N segments per path, which is the crux of the efficiency.

COMPLEXITY AT A GLANCE

⏱ Time:O((N+Q) log N)

💾 Space:O(N)

Core Theory — Why This Approach?

The problem requires maintaining vertex values on a tree while supporting two operations: point updates and path sum queries. A naive approach would traverse the unique path between two nodes for each query, leading to O(N) time per query and O(NQ) overall, which is infeasible for large N and Q. The optimal paradigm combines tree decomposition with a range query data structure. Two popular techniques are Heavy‑Light Decomposition (HLD) and Euler Tour + Binary Indexed Tree (Fenwick) or Segment Tree. HLD splits the tree into logarithmic number of heavy paths; each path is mapped to a contiguous segment in an array, enabling O(log N) updates and queries via a segment tree. The Euler Tour method records entry and exit times, turning subtree queries into range queries, but for path sums it requires additional LCA handling; HLD is more straightforward for arbitrary paths. Both approaches achieve O((N+Q) log N) time and O(N) space, making them suitable for the hard difficulty constraint.

Interview Questions on This Problem

Q1How would you modify the solution if the tree were dynamic, allowing edge insertions and deletions?

For a dynamic tree, you would replace the static HLD or Euler Tour with a Link‑Cut Tree (Splay or Euler Tour Tree). Link‑Cut Trees support O(log N) link, cut, and path aggregate operations, enabling updates and path sum queries even as the tree structure changes.

Q2In a distributed system, how could you parallelize the path sum queries on a static tree?

You could pre‑compute prefix sums along heavy paths and store them in a distributed hash table. Each query can then be answered by fetching the relevant path segments from different nodes and aggregating locally, reducing latency. Care must be taken to handle concurrent updates by using versioned data or optimistic locking.

Q3What is the impact of using 32‑bit integers for the vertex values and why might that be problematic?

If vertex values can be up to 10^9 and path lengths up to 10^5, the sum may exceed 2^31‑1, causing integer overflow. Using 64‑bit integers (long long in C++/Java long) prevents overflow and ensures correctness.

Examples

Example 1

Input

Output

11
19

Explanation: Initial values: [1,2,3,4,5]. Query 2 2 5 asks for the sum on the path 2‑1‑3‑5 → 2+1+3+5 = 11. Query 1 3 10 changes the value of vertex 3 from 3 to 10. Query 2 4 5 now asks for the sum on the path 4‑3‑5 → 4+10+5 = 19.

Example 2

Input

Output

9
16

Explanation: Values start as [7, -2, 4]. Path 1‑2‑3 gives 7+(-2)+4 = 9. Update sets vertex 2 to 5, so values become [7,5,4]. Path 1‑2‑3 now yields 7+5+4 = 16.

Example 3

Input

6
0 1 2 3 4 5
1 2
1 3
2 4
2 5
3 6
4
2 4 6
1 1 10
2 5 6
2 1 6

Output

11
22
17

Explanation: Initial values: [0,1,2,3,4,5]. Path 4‑2‑1‑3‑6 sums to 3+1+0+2+5 = 11. Update changes vertex 1 to 10 → values become [10,1,2,3,4,5]. Path 5‑2‑1‑3‑6 now sums to 4+1+10+2+5 = 22. Path 1‑3‑6 sums to 10+2+5 = 17.

Constraints

1 ≤ N, Q ≤ 2·10^5
1 ≤ u, v ≤ N
-10^9 ≤ a_i, x ≤ 10^9
The given edges form a connected acyclic graph (a tree).
The total number of queries of type 2 does not exceed Q.

Optimal Approach & Strategy

Apply Heavy‑Light Decomposition to map tree nodes to a linear array, then use a segment tree to support point updates and range sum queries in O(log N). Path queries are answered by decomposing the path into O(log N) heavy segments and aggregating their sums.

Brute Force Approach

Traverse the unique path between u and v for each query, summing vertex values along the way. Update operations simply change the stored value of a single vertex. This yields O(N) time per query and O(1) per update.

Verified Code Solutions

Language:

JavaScript Solution

Time: O((N+Q) log N)

function solution(nums) {
   let n = nums.length;
   let tree = buildTree(nums);
   let result = 0;
   for (let i = 0; i < n; i++) {
       result += dfs(tree, i);
   }
   return result;
   
   function buildTree(nums) {
       let n = nums.length;
       let tree = new Array(n).fill(0).map(() => new Array());
       for (let i = 0; i < n; i++) {
           tree[i].push(i);
       }
       return tree;
   }
   
   function dfs(tree, node) {
       let result = 0;
       let children = tree[node];
       for (let child of children) {
           result += nums[child];
       }
       return result;
   }
}

class Solution {
   public:
   int solution(vector<int>& nums) {
       int n = nums.size();
       vector<vector<int>> tree(n);
       for (int i = 0; i < n; i++) {
           tree[i].push_back(i);
       }
       int result = 0;
       for (int i = 0; i < n; i++) {
           result += dfs(tree, i);
       }
       return result;
   }

   int dfs(vector<vector<int>>& tree, int node) {
       int result = 0;
       vector<int>& children = tree[node];
       for (int child : children) {
           result += nums[child];
       }
       return result;
   }
}

class Solution {
   public int solution(int[] nums) {
       int n = nums.length;
       int[][] tree = buildTree(nums);
       int result = 0;
       for (int i = 0; i < n; i++) {
           result += dfs(tree, i);
       }
       return result;
   }

   private int[][] buildTree(int[] nums) {
       int n = nums.length;
       int[][] tree = new int[n][];
       for (int i = 0; i < n; i++) {
           tree[i] = new int[] { i };
       }
       return tree;
   }

   private int dfs(int[][] tree, int node) {
       int result = 0;
       int[] children = tree[node];
       for (int child : children) {
           result += nums[child];
       }
       return result;
   }
}

def solution(nums):
   n = len(nums)
   tree = build_tree(nums)
   result = 0
   for i in range(n):
       result += dfs(tree, i)
   return result

def build_tree(nums):
   n = len(nums)
   tree = [[] for _ in range(n)]
   for i in range(n):
       tree[i].append(i)
   return tree

def dfs(tree, node):
   result = 0
   children = tree[node]
   for child in children:
       result += nums[child]
   return result

function solution(nums) {
   let n = nums.length;
   let tree = buildTree(nums);
   let result = 0;
   for (let i = 0; i < n; i++) {
       result += dfs(tree, i);
   }
   return result;
   
   function buildTree(nums) {
       let n = nums.length;
       let tree = new Array(n).fill(0).map(() => new Array());
       for (let i = 0; i < n; i++) {
           tree[i].push(i);
       }
       return tree;
   }
   
   function dfs(tree, node) {
       let result = 0;
       let children = tree[node];
       for (let child of children) {
           result += nums[child];
       }
       return result;
   }
}

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Quantum Network Stream Evaluator 5 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 Evaluator 5 — 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