Quantum Network Stream Evaluator — Problem Statement & Solution Guide
Quick Answer & Algorithm Key Takeaway
Master coding challenges related to Graphs and solve the Quantum Network Stream Evaluator 4 problem optimally.
O(N)O(N)Problem Description
You are tasked with analyzing the structural balance of a quantum communication network modeled as a tree with N nodes. The network's stability is determined by its centroid: a node whose removal minimizes the size of the largest resulting connected component. Specifically, a node u is a centroid if the maximum size of any subtree rooted at a neighbor of u (considering u as the root of the entire tree) is at most N/2. Your goal is to identify all such centroid nodes in the given tree.
To solve this efficiently for large networks, you must implement the Heavy-Light Decomposition (HLD) technique. Although HLD is typically used for path queries, its underlying decomposition into heavy and light paths provides a natural framework for efficiently computing subtree sizes and identifying the 'heavy' child (the child with the largest subtree) at each node. By leveraging the HLD structure, you can traverse the tree to compute the maximum component size for each node in O(N) time after an O(N) preprocessing step, which is optimal for this problem.
Input: An integer N representing the number of nodes, followed by N-1 lines, each containing two integers u and v, indicating an undirected edge between nodes u and v. Nodes are 1-indexed.
Output: A list of all centroid node indices, sorted in ascending order. If multiple centroids exist, return all of them.
DSA Pattern Breakdown
DSA Pattern Breakdown
"Quantum Network Stream Evaluator"
WHY DOES IT MATTER?
Centroid detection is a fundamental tree‑balancing pattern that underpins many advanced techniques like centroid decomposition, heavy‑light decomposition, and optimal root selection for minimizing height or depth. Mastery of this pattern unlocks efficient solutions for a wide class of tree queries.
OPTIMIZATION CHALLENGE
The breakthrough is realizing that the size of the component containing the parent can be derived from the total node count and the already‑computed subtree size, eliminating the need for a second full traversal per node.
REAL-WORLD CONNECTION
Think of a corporate hierarchy where you need to appoint a manager whose removal would not leave any department larger than half the company. Choosing the centroid ensures that any re‑organization after the manager leaves keeps departments balanced, mirroring load‑balancing in distributed databases.
During an interview, compute subtree sizes in one DFS, then reuse those values in a second pass (or the same pass) to evaluate the balance condition; avoid recomputing sizes for each neighbor—this keeps the solution O(N).
COMPLEXITY AT A GLANCE
O(N)O(N)Core Theory — Why This Approach?
In a tree, a centroid is a node whose removal leaves all resulting connected components with size at most N/2. The classic linear‑time algorithm computes subtree sizes via a single DFS, then for each node evaluates the largest component formed by its parent side (N - subtreeSize[node]) and each child side (subtreeSize[child]). The node is a centroid if the maximum of these values does not exceed N/2. Naïve approaches that recompute component sizes after removing each node would require O(N^2) time because each removal triggers a fresh traversal, which is infeasible for N up to 2·10^5 or higher. The optimal paradigm leverages the tree’s hierarchical structure: a single post‑order pass gathers subtree sizes, and a second pass (or the same pass with careful bookkeeping) checks the balance condition for every node in O(1) per node, yielding overall O(N) time and O(N) auxiliary space.
Interview Questions on This Problem
Q1How would you find all centroids of a tree in linear time and why can a tree have at most two centroids?
Perform one DFS to compute subtree sizes. For each node, compute the size of the largest component after its removal as max(N - subtreeSize[node], max{subtreeSize[child] for each child}). Nodes where this value ≤ N/2 are centroids. A tree can have at most two centroids because if there were three, the middle one would split the tree into three parts each ≤ N/2, forcing the total size > N.
Q2Explain how the centroid concept can be used to design a divide‑and‑conquer algorithm on trees (e.g., centroid decomposition).
Centroid decomposition recursively selects a centroid, removes it, and processes each resulting subtree independently. Since each removal guarantees that every subtree has size ≤ N/2, the depth of recursion is O(log N). This property enables efficient solutions for path queries, distance calculations, and dynamic programming on trees.
Q3In a distributed system modeled as a tree, why might you choose a centroid node as a leader for fault tolerance?
A centroid minimizes the size of the largest partition after failure, ensuring that no single failure isolates a majority of nodes. This balances load and reduces worst‑case communication latency, making the system more resilient to node crashes.
Examples
Input
N = 5 Edges: [[1,2], [2,3], [2,4], [2,5]]
Output
[2]
Explanation: The tree is a star-like structure centered at node 2. Removing node 2 results in four components of size 1 each. The maximum component size is 1, which is <= 5/2 = 2.5. For any other node (e.g., node 1), removing it leaves a component of size 4 (nodes 2,3,4,5), which is > 2.5. Thus, only node 2 is a centroid.
Input
N = 6 Edges: [[1,2], [2,3], [3,4], [4,5], [5,6]]
Output
[3, 4]
Explanation: This is a linear chain. Removing node 3 splits the tree into a component of size 2 (nodes 1,2) and a component of size 3 (nodes 4,5,6). Max size is 3, which is <= 6/2 = 3. Removing node 4 splits it into size 3 (nodes 1,2,3) and size 2 (nodes 5,6). Max size is 3, which is <= 3. Nodes 2 and 5 have max component sizes of 4 and 4 respectively, which are > 3. Thus, centroids are 3 and 4.
Input
N = 7 Edges: [[1,2], [1,3], [2,4], [2,5], [3,6], [3,7]]
Output
[1]
Explanation: Node 1 is the root of a balanced binary tree. Removing node 1 results in two components: one rooted at 2 (size 3: nodes 2,4,5) and one rooted at 3 (size 3: nodes 3,6,7). Max size is 3, which is <= 7/2 = 3.5. For node 2, removing it leaves a component containing node 1 and its other subtree (size 4: nodes 1,3,6,7) and two single nodes. Max size is 4 > 3.5. Similarly for node 3. Thus, only node 1 is a centroid.
Constraints
- 2 <= N <= 10^5
- 1 <= u, v <= N
- The input graph is guaranteed to be a valid tree (connected and acyclic).
- The sum of N over all test cases does not exceed 10^6.
Optimal Approach & Strategy
Compute subtree sizes with a single DFS, then for each node evaluate max(N - subtreeSize[node], max child subtree sizes) in O(1) per node, yielding O(N) total.
Brute Force Approach
Remove each node one by one, run a BFS/DFS to compute the size of every resulting component, and keep nodes where the largest component ≤ N/2. This costs O(N^2).
Code Solutions
function solution(tree) {
let centroid = -1;
let maxSum = -1;
for (let i = 0; i < tree.length; i++) {
let sum = 0;
let stack = [i];
let visited = new Set();
while (stack.length > 0) {
let node = stack.pop();
if (!visited.has(node)) {
sum += tree[node];
visited.add(node);
for (let j = 0; j < tree.length; j++) {
if (tree[j] !== 0 && !visited.has(j)) {
stack.push(j);
}
}
}
}
if (sum > maxSum) {
maxSum = sum;
centroid = i;
}
}
return centroid;
}class Solution {
public:
int solution(vector<int>& tree) {
int centroid = -1;
int maxSum = -1;
for (int i = 0; i < tree.size(); i++) {
int sum = 0;
vector<bool> visited(tree.size(), false);
sum = dfs(tree, i, visited);
if (sum > maxSum) {
maxSum = sum;
centroid = i;
}
}
return centroid;
}
int dfs(vector<int>& tree, int node, vector<bool>& visited) {
if (visited[node]) return 0;
visited[node] = true;
int sum = tree[node];
for (int i = 0; i < tree.size(); i++) {
if (tree[i] != 0 && !visited[i]) {
sum += dfs(tree, i, visited);
}
}
return sum;
}
}class Solution {
public int solution(int[] tree) {
int centroid = -1;
int maxSum = -1;
for (int i = 0; i < tree.length; i++) {
int sum = 0;
boolean[] visited = new boolean[tree.length];
sum = dfs(tree, i, visited);
if (sum > maxSum) {
maxSum = sum;
centroid = i;
}
}
return centroid;
}
public int dfs(int[] tree, int node, boolean[] visited) {
if (visited[node]) return 0;
visited[node] = true;
int sum = tree[node];
for (int i = 0; i < tree.length; i++) {
if (tree[i] != 0 && !visited[i]) {
sum += dfs(tree, i, visited);
}
}
return sum;
}
}def solution(tree):
centroid = -1
max_sum = -1
for i in range(len(tree)):
sum = 0
stack = [i]
visited = set()
while stack:
node = stack.pop()
if node not in visited:
sum += tree[node]
visited.add(node)
for j in range(len(tree)):
if tree[j] != 0 and j not in visited:
stack.append(j)
if sum > max_sum:
max_sum = sum
centroid = i
return centroidfunction solution(tree) {
let centroid = -1;
let maxSum = -1;
for (let i = 0; i < tree.length; i++) {
let sum = 0;
let stack = [i];
let visited = new Set();
while (stack.length > 0) {
let node = stack.pop();
if (!visited.has(node)) {
sum += tree[node];
visited.add(node);
for (let j = 0; j < tree.length; j++) {
if (tree[j] !== 0 && !visited.has(j)) {
stack.push(j);
}
}
}
}
if (sum > maxSum) {
maxSum = sum;
centroid = i;
}
}
return centroid;
}Asked in Top Tech Interviews
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.