Count Number of Possible Root Nodes solution and explanation
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The problem
Alice has an undirected tree with n nodes labeled from 0 to n - 1. The tree is represented as a 2D integer array edges of length n - 1 where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.
Alice wants Bob to find the root of the tree. She allows Bob to make several guesses about her tree. In one guess, he does the following:
- Chooses two distinct integers
uandvsuch that there exists an edge[u, v]in the tree. - He tells Alice that
uis the parent ofvin the tree.
Bob's guesses are represented by a 2D integer array guesses where guesses[j] = [uj, vj] indicates Bob guessed uj to be the parent of vj.
Alice being lazy, does not reply to each of Bob's guesses, but just says that at least k of his guesses are true.
Given the 2D integer arrays edges, guesses and the integer k, return the number of possible nodes that can be the root of Alice's tree. If there is no such tree, return 0.
Example 1:
Input: edges = [[0,1],[1,2],[1,3],[4,2]], guesses = [[1,3],[0,1],[1,0],[2,4]], k = 3 Output: 3 Explanation: Root = 0, correct guesses = [1,3], [0,1], [2,4] Root = 1, correct guesses = [1,3], [1,0], [2,4] Root = 2, correct guesses = [1,3], [1,0], [2,4] Root = 3, correct guesses = [1,0], [2,4] Root = 4, correct guesses = [1,3], [1,0] Considering 0, 1, or 2 as root node leads to 3 correct guesses.
Example 2:
Input: edges = [[0,1],[1,2],[2,3],[3,4]], guesses = [[1,0],[3,4],[2,1],[3,2]], k = 1 Output: 5 Explanation: Root = 0, correct guesses = [3,4] Root = 1, correct guesses = [1,0], [3,4] Root = 2, correct guesses = [1,0], [2,1], [3,4] Root = 3, correct guesses = [1,0], [2,1], [3,2], [3,4] Root = 4, correct guesses = [1,0], [2,1], [3,2] Considering any node as root will give at least 1 correct guess.
Constraints:
edges.length == n - 12 <= n <= 1051 <= guesses.length <= 1050 <= ai, bi, uj, vj <= n - 1ai != biuj != vjedgesrepresents a valid tree.guesses[j]is an edge of the tree.guessesis unique.0 <= k <= guesses.length
Approach overview
Problem Overview: You are given an undirected tree with n nodes and a list of parent-child guesses. A guess (u, v) means someone believes u is the parent of v. For a chosen root, the tree becomes directed from parent to child. The task is to count how many nodes could be valid roots such that at least k guesses become correct.
Approach 1: Depth-First Search with Rerooting (O(n))
Store all guesses in a HashSet for constant-time lookup. First root the tree at node 0 and run a Depth-First Search to count how many guesses match the parent-child direction. Each DFS step checks whether (parent, child) exists in the guess set. This gives the number of correct guesses if the root is 0. Next, reroot the tree using a second DFS: when moving the root from u to v, adjust the score by subtracting (u, v) if it was a correct guess and adding (v, u) if that guess exists. This rerooting trick reuses previous computation instead of recomputing from scratch.
The key insight: changing the root only affects the orientation of one edge. Maintaining the score incrementally avoids repeating full traversals. Using adjacency lists for the tree and a hash table for guesses keeps operations constant time. Total complexity is O(n) time and O(n) space.
Approach 2: Union-Find (Disjoint Set Union) Assisted Validation (O(n α(n)))
Union-Find can be used to manage connectivity while evaluating potential parent-child relationships implied by guesses. Build the tree using adjacency lists, then use DSU to ensure edges belong to the same connected component and to quickly verify relationships when testing possible roots. For each node treated as root, validate guess directions along the tree structure while relying on DSU to avoid repeated connectivity checks. This reduces structural overhead but still requires traversing edges for orientation validation.
This method runs in roughly O(n α(n)) time due to near-constant amortized union/find operations and uses O(n) space. It is less common for interviews but can simplify connectivity reasoning when combining multiple structural constraints.
Recommended for interviews: The DFS rerooting approach is the expected solution. Start by computing correct guesses for a fixed root, then apply reroot dynamic programming to update counts in constant time per edge. This demonstrates strong understanding of tree rerooting DP and efficient graph traversal. A brute-force attempt (testing every node as root with full DFS) shows baseline reasoning but runs in O(n²), which is too slow for large inputs.
Approach 1: Depth-First Search (DFS) with Rooted Tree
This approach involves building the tree and performing a Depth-First Search (DFS) from an arbitrary node, usually node 0, to determine the initial configuration or parent-child relationships. By maintaining awareness of which guesses match the found parent-child relationships, you can compute the number of valid roots by rerooting the tree at each node.
from collections import defaultdict
class Solution:
def possibleRootCount(self, edges, guesses, k):
n = len(edges) + 1
tree = defaultdict(list)
for u, v in edges:
tree[u].append(v)
tree[v].append(u)
guesses_set = {(u, v) for u, v in guesses}
valid_roots = [0] * n
def dfs(node, parent, correct):
if (parent, node) in guesses_set:
correct += 1
for neighbor in tree[node]:
if neighbor != parent:
correct = dfs(neighbor, node, correct)
return correct
def reroot_and_count(node, parent, correct):
if correct >= k:
valid_roots[node] = 1
for neighbor in tree[node]:
if neighbor != parent:
# root shift
if (node, neighbor) in guesses_set:
correct -= 1
if (neighbor, node) in guesses_set:
correct += 1
reroot_and_count(neighbor, node, correct)
# restore root
if (node, neighbor) in guesses_set:
correct += 1
if (neighbor, node) in guesses_set:
correct -= 1
correct_guesses = dfs(0, -1, 0)
reroot_and_count(0, -1, correct_guesses)
return sum(valid_roots)
# Sample usage
egdes = [[0, 1], [1, 2], [1, 3], [4, 2]]
guesses = [[1, 3], [0, 1], [1, 0], [2, 4]]
k = 3
sol = Solution()
print(sol.possibleRootCount(edges, guesses, k)) # Output: 3import java.util.*;
class Solution {
public int possibleRootCount(int[][] edges, int[][] guesses, int k) {
int n = edges.length + 1;
List<Integer>[] tree = new ArrayList[n];
for (int i = 0; i < n; i++) tree[i] = new ArrayList<>();
Set<Pair<Integer, Integer>> guessSet = new HashSet<>();
for (int[] edge : edges) {
tree[edge[0]].add(edge[1]);
tree[edge[1]].add(edge[0]);
}
for (int[] guess : guesses) {
guessSet.add(new Pair<>(guess[0], guess[1]));
}
int[] validRoots = new int[n];
int correct = 0;
dfs(tree, 0, -1, guessSet, correct);
reroot(tree, 0, -1, guessSet, validRoots, correct, k);
int count = 0;
for (int v : validRoots) if (v == 1) count++;
return count;
}
private void dfs(List<Integer>[] tree, int node, int parent, Set<Pair<Integer, Integer>> guessSet, int correct) {
for (int child : tree[node]) {
if (child != parent) {
Pair<Integer, Integer> guessPair = new Pair<>(node, child);
if (guessSet.contains(guessPair)) correct++;
dfs(tree, child, node, guessSet, correct);
}
}
}
private void reroot(List<Integer>[] tree, int node, int parent, Set<Pair<Integer, Integer>> guessSet, int[] validRoots, int correct, int k) {
if (correct >= k) validRoots[node] = 1;
for (int child: tree[node]) {
if (child != parent) {
Pair<Integer, Integer> guessPairCur = new Pair<>(node, child);
Pair<Integer, Integer> guessPairRev = new Pair<>(child, node);
if (guessSet.contains(guessPairCur)) correct--;
if (guessSet.contains(guessPairRev)) correct++;
reroot(tree, child, node, guessSet, validRoots, correct, k);
if (guessSet.contains(guessPairCur)) correct++;
if (guessSet.contains(guessPairRev)) correct--;
}
}
}
static class Pair<K, V> {
K key;
V value;
Pair(K key, V value) {
this.key = key;
this.value = value;
}
@Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pair<?, ?> pair = (Pair<?, ?>) o;
return Objects.equals(key, pair.key) && Objects.equals(value, pair.value);
}
@Override
public int hashCode() {
return Objects.hash(key, value);
}
}
public static void main(String[] args) {
Solution sol = new Solution();
int[][] edges = {{0, 1}, {1, 2}, {1, 3}, {4, 2}};
int[][] guesses = {{1, 3}, {0, 1}, {1, 0}, {2, 4}};
int k = 3;
System.out.println(sol.possibleRootCount(edges, guesses, k));
}
}#include <iostream>
#include <vector>
#include <unordered_map>
#include <unordered_set>
using namespace std;
class Solution {
public:
int possibleRootCount(vector<vector<int>>& edges, vector<vector<int>>& guesses, int k) {
int n = edges.size() + 1;
vector<vector<int>> tree(n);
unordered_map<int, unordered_set<int>> guessMap;
for (auto& edge : edges) {
tree[edge[0]].push_back(edge[1]);
tree[edge[1]].push_back(edge[0]);
}
for (auto& guess : guesses) {
guessMap[guess[0]].insert(guess[1]);
}
int correct = 0;
vector<int> validRoots(n, 0);
dfs(tree, 0, -1, guessMap, correct);
reroot(tree, 0, -1, guessMap, validRoots, correct, k);
int count = 0;
for (int v : validRoots) if (v) ++count;
return count;
}
private:
void dfs(vector<vector<int>>& tree, int node, int parent, unordered_map<int, unordered_set<int>>& guessMap, int& correct) {
for (int child : tree[node]) {
if (child != parent) {
if (guessMap[node].count(child)) ++correct;
dfs(tree, child, node, guessMap, correct);
}
}
}
void reroot(vector<vector<int>>& tree, int node, int parent, unordered_map<int, unordered_set<int>>& guessMap, vector<int>& validRoots, int& correct, int k) {
if (correct >= k) validRoots[node] = 1;
for (int child: tree[node]) {
if (child != parent) {
if (guessMap[node].count(child)) correct--;
if (guessMap[child].count(node)) correct++;
reroot(tree, child, node, guessMap, validRoots, correct, k);
if (guessMap[node].count(child)) correct++;
if (guessMap[child].count(node)) correct--;
}
}
}
};
int main() {
vector<vector<int>> edges = {{0,1},{1,2},{1,3},{4,2}};
vector<vector<int>> guesses = {{1,3},{0,1},{1,0},{2,4}};
int k = 3;
Solution sol;
cout << sol.possibleRootCount(edges, guesses, k) << endl;
return 0;
}class Solution {
possibleRootCount(edges, guesses, k) {
const n = edges.length + 1;
const tree = Array.from({ length: n }, () => []);
const guessSet = new Set();
for (const [u, v] of edges) {
tree[u].push(v);
tree[v].push(u);
}
for (const [u, v] of guesses) {
guessSet.add(`${u},${v}`);
}
let correct = 0;
const validRoots = Array(n).fill(0);
this.dfs(tree, 0, -1, guessSet, correct);
this.reroot(tree, 0, -1, guessSet, validRoots, correct, k);
return validRoots.reduce((acc, val) => acc + val, 0);
}
dfs(tree, node, parent, guessSet, correct) {
for (const child of tree[node]) {
if (child !== parent) {
if (guessSet.has(`${node},${child}`)) correct++;
this.dfs(tree, child, node, guessSet, correct);
}
}
}
reroot(tree, node, parent, guessSet, validRoots, correct, k) {
if (correct >= k) validRoots[node] = 1;
for (const child of tree[node]) {
if (child !== parent) {
if (guessSet.has(`${node},${child}`)) correct--;
if (guessSet.has(`${child},${node}`)) correct++;
this.reroot(tree, child, node, guessSet, validRoots, correct, k);
if (guessSet.has(`${node},${child}`)) correct++;
if (guessSet.has(`${child},${node}`)) correct--;
}
}
}
}
// Usage
const edges = [[0, 1], [1, 2], [1, 3], [4, 2]];
const guesses = [[1, 3], [0, 1], [1, 0], [2, 4]];
const k = 3;
const sol = new Solution();
console.log(sol.possibleRootCount(edges, guesses, k));#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define MAX_NODES 100000
int graph[MAX_NODES][MAX_NODES];
int visited[MAX_NODES];
int guessMap[MAX_NODES][MAX_NODES];
int guessCount = 0;
int match[MAX_NODES];
void dfs(int node, int parent) {
visited[node] = 1;
for (int i = 0; i < MAX_NODES; ++i) {
if (graph[node][i] && i != parent) {
if (guessMap[node][i] == 1) {
guessCount++;
}
dfs(i, node);
}
}
}
int main() {
int edges[][2] = {{0,1},{1,2},{1,3},{4,2}};
int guesses[][2] = {{1,3},{0,1},{1,0},{2,4}};
int k = 3;
int n = 5;
memset(graph, 0, sizeof(graph));
memset(visited, 0, sizeof(visited));
memset(guessMap, 0, sizeof(guessMap));
// Building the graph
for (int i = 0; i < n - 1; ++i) {
int u = edges[i][0];
int v = edges[i][1];
graph[u][v] = 1;
graph[v][u] = 1;
}
// Building the guess map
for (int i = 0; i < sizeof(guesses) / sizeof(guesses[0]); i++) {
int u = guesses[i][0];
int v = guesses[i][1];
guessMap[u][v] = 1;
}
int possibleRoots = 0;
for (int i = 0; i < n; ++i) {
memset(visited, 0, sizeof(visited));
guessCount = 0;
dfs(i, -1);
if (guessCount >= k) {
possibleRoots++;
}
}
printf("%d\n", possibleRoots);
return 0;
}using System;
using System.Collections.Generic;
public class Solution {
public int PossibleRootCount(int[][] edges, int[][] guesses, int k) {
int n = edges.Length + 1;
List<int>[] tree = new List<int>[n];
for (int i = 0; i < n; i++) tree[i] = new List<int>();
HashSet<(int, int)> guessSet = new HashSet<(int, int)>();
foreach (var edge in edges) {
tree[edge[0]].Add(edge[1]);
tree[edge[1]].Add(edge[0]);
}
foreach (var guess in guesses) {
guessSet.Add((guess[0], guess[1]));
}
int[] validRoots = new int[n];
int correct = 0;
Dfs(tree, 0, -1, guessSet, ref correct);
Reroot(tree, 0, -1, guessSet, validRoots, ref correct, k);
int count = 0;
foreach (int v in validRoots) if (v == 1) count++;
return count;
}
private void Dfs(List<int>[] tree, int node, int parent, HashSet<(int, int)> guessSet, ref int correct) {
foreach (int child in tree[node]) {
if (child != parent) {
if (guessSet.Contains((node, child))) correct++;
Dfs(tree, child, node, guessSet, ref correct);
}
}
}
private void Reroot(List<int>[] tree, int node, int parent, HashSet<(int, int)> guessSet, int[] validRoots, ref int correct, int k) {
if (correct >= k) validRoots[node] = 1;
foreach (int child in tree[node]) {
if (child != parent) {
if (guessSet.Contains((node, child))) correct--;
if (guessSet.Contains((child, node))) correct++;
Reroot(tree, child, node, guessSet, validRoots, ref correct, k);
if (guessSet.Contains((node, child))) correct++;
if (guessSet.Contains((child, node))) correct--;
}
}
}
public static void Main(string[] args) {
var edges = new int[][] { new int[]{0, 1}, new int[]{1, 2}, new int[]{1, 3}, new int[]{4, 2} };
var guesses = new int[][] { new int[]{1, 3}, new int[]{0, 1}, new int[]{1, 0}, new int[]{2, 4} };
int k = 3;
Solution sol = new Solution();
Console.WriteLine(sol.PossibleRootCount(edges, guesses, k));
}
}In this Python approach, a defaultdict is used to implement adjacency lists, allowing dynamic expansion. A set is used to store guesses for O(1) retrieval time. Two functions, dfs (for calculating initial guesses) and reroot_and_count (for adjusting guesses for each node as the root), provide a recursive framework.
Time Complexity: O(n), as the adjacency list optimizes node traversal, with set lookups ensuring constant time for guesses.
Space Complexity: O(n + g), given the space-efficient structure for storing tree and guesses.
Approach 2: Using Union-Find (Disjoint Set Union, DSU)
This approach utilizes the Union-Find data structure to keep track of connected components in an efficient way—each node as its own separate component and merging them via edges. By adapting the structure for parent relationships and examining which parent associations hold true, we can determine potential roots that meet the guessing criteria.
class UnionFind:
def __init__(self, size):
self.root = [i for i in range(size)]
self.rank = [1] * size
def find(self, x):
if self.root[x] == x:
return x
self.root[x] = self.find(self.root[x])
return self.root[x]
def union(self, x, y):
rootX = self.find(x)
rootY = self.find(y)
if rootX != rootY:
if self.rank[rootX] > self.rank[rootY]:
self.root[rootY] = rootX
elif self.rank[rootX] < self.rank[rootY]:
self.root[rootX] = rootY
else:
self.root[rootY] = rootX
self.rank[rootX] += 1
class Solution:
def possibleRootCount(self, edges, guesses, k):
n = len(edges) + 1
uf = UnionFind(n)
for u, v in edges:
uf.union(u, v)
correct_count = {}
for u, v in guesses:
root_u = uf.find(u)
if uf.find(v) == root_u:
correct_count[root_u] = correct_count.get(root_u, 0) + 1
return sum(1 for node in range(n) if correct_count.get(uf.find(node), 0) >= k)
# Usage
sol = Solution()
edges = [[0, 1], [1, 2], [1, 3], [4, 2]]
guesses = [[1, 3], [0, 1], [1, 0], [2, 4]]
k = 3
print(sol.possibleRootCount(edges, guesses, k)) # Output: 3#include <iostream>
#include <vector>
#include <unordered_map>
#include <unordered_set>
using namespace std;
class Solution {
public:
int findRoot(vector<int>& parent, int node) {
if (parent[node] == node) return node;
return parent[node] = findRoot(parent, parent[node]);
}
void unionSets(vector<int>& parent, vector<int>& rank, int u, int v) {
int rootU = findRoot(parent, u);
int rootV = findRoot(parent, v);
if (rootU != rootV) {
if (rank[rootU] > rank[rootV]) parent[rootV] = rootU;
else if (rank[rootU] < rank[rootV]) parent[rootU] = rootV;
else { parent[rootV] = rootU; rank[rootU]++; }
}
}
int possibleRootCount(vector<vector<int>>& edges, vector<vector<int>>& guesses, int k) {
int n = edges.size() + 1;
vector<int> parent(n), rank(n, 0), validRoots(n, 0);
for (int i = 0; i < n; ++i) parent[i] = i;
for (const auto& edge: edges) unionSets(parent, rank, edge[0], edge[1]);
unordered_map<int, int> correctCount;
unordered_set<int> guessSet;
for (const auto& guess : guesses) {
int p = findRoot(parent, guess[0]);
if (findRoot(parent, guess[1]) == p) correctCount[p]++;
}
int count = 0;
for (int i = 0; i < n; ++i) {
if (correctCount[findRoot(parent, i)] >= k) count ++;
}
return count;
}
};
int main() {
Solution sol;
vector<vector<int>> edges = {{0,1},{1,2},{1,3},{4,2}};
vector<vector<int>> guesses = {{1,3},{0,1},{1,0},{2,4}};
int k = 3;
cout << sol.possibleRootCount(edges, guesses, k) << endl;
return 0;
}In the Union-Find approach in Python, two arrays manage the parent/root and rank of each node/component. After merging, a mapping of roots to correct guesses calculates correctness for potential roots. Traversing each node and comparing correct guesses to an input threshold determines the validity of roots.
Time Complexity: O(n*α(n)), with α being the inverse Ackermann function due to use of path compression and union by rank, which are efficient operations.
Space Complexity: O(n), as both array structures size up to the node count.
Approach 3: Tree DP (change root)
First, we traverse the given edge set edges and convert it to an adjacency list g where g[i] represents the adjacent nodes of node i. Use a hash map gs to record the given guess set guesses.
Then, we start from node 0 and perform a DFS to count the number of edges in guesses among all the nodes that can be reached from node 0. We use the variable cnt to record this number.
Next, we start from node 0 and perform a DFS to count the number of edges in guesses in each tree with 0 as the root. If the number is greater than or equal to k, it means that this node is a possible root node, and we add 1 to the answer.
Therefore, the problem becomes to count the number of edges in guesses in each tree with each node as the root. We already know that there are cnt edges in guesses among all the nodes that can be reached from node 0. We can maintain this value by adding or subtracting the current edge in guesses in DFS.
Assume that we are currently traversing node i and cnt represents the number of edges in guesses with i as the root node. Then, for each adjacent node j of i, we need to calculate the number of edges in guesses with j as the root node. If (i, j) is in guesses, then there is no edge (i, j) in the tree with j as the root node, so cnt should decrease by 1. If (j, i) is in guesses, then there is an extra edge (i, j) in the tree with j as the root node, so cnt should increase by 1. That is, f[j] = f[i] + (j, i) \in guesses - (i, j) \in guesses. Where f[i] represents the number of edges in guesses with i as the root node.
The time complexity is O(n + m) and the space complexity is O(n + m), where n and m are the lengths of edges and guesses respectively.
function rootCount(edges: number[][], guesses: number[][], k: number): number {
const n = edges.length + 1;
const g: number[][] = Array.from({ length: n }, () => []);
const gs: Map<number, number> = new Map();
const f = (i: number, j: number) => i * n + j;
for (const [a, b] of edges) {
g[a].push(b);
g[b].push(a);
}
for (const [a, b] of guesses) {
const x = f(a, b);
gs.set(x, gs.has(x) ? gs.get(x)! + 1 : 1);
}
let ans = 0;
let cnt = 0;
const dfs1 = (i: number, fa: number): void => {
for (const j of g[i]) {
if (j !== fa) {
cnt += gs.get(f(i, j)) || 0;
dfs1(j, i);
}
}
};
const dfs2 = (i: number, fa: number): void => {
ans += cnt >= k ? 1 : 0;
for (const j of g[i]) {
if (j !== fa) {
const a = gs.get(f(i, j)) || 0;
const b = gs.get(f(j, i)) || 0;
cnt -= a;
cnt += b;
dfs2(j, i);
cnt -= b;
cnt += a;
}
}
};
dfs1(0, -1);
dfs2(0, -1);
return ans;
}func rootCount(edges [][]int, guesses [][]int, k int) (ans int) {
n := len(edges) + 1
g := make([][]int, n)
gs := map[int]int{}
for _, e := range edges {
a, b := e[0], e[1]
g[a] = append(g[a], b)
g[b] = append(g[b], a)
}
f := func(i, j int) int {
return i*n + j
}
for _, e := range guesses {
a, b := e[0], e[1]
gs[f(a, b)]++
}
cnt := 0
var dfs1 func(i, fa int)
var dfs2 func(i, fa int)
dfs1 = func(i, fa int) {
for _, j := range g[i] {
if j != fa {
cnt += gs[f(i, j)]
dfs1(j, i)
}
}
}
dfs2 = func(i, fa int) {
if cnt >= k {
ans++
}
for _, j := range g[i] {
if j != fa {
a, b := gs[f(i, j)], gs[f(j, i)]
cnt -= a
cnt += b
dfs2(j, i)
cnt -= b
cnt += a
}
}
}
dfs1(0, -1)
dfs2(0, -1)
return
}Complexity comparison
| Approach | Complexity |
|---|---|
| Depth-First Search (DFS) with Rooted Tree | Time Complexity: O(n), as the adjacency list optimizes node traversal, with set lookups ensuring constant time for guesses. |
| Using Union-Find (Disjoint Set Union, DSU) | Time Complexity: O(n*α(n)), with α being the inverse Ackermann function due to use of path compression and union by rank, which are efficient operations. |
| Tree DP (change root) | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Brute Force: Try Every Node as Root with DFS | O(n²) | O(n) | Conceptual baseline or very small trees |
| DFS with Rerooting and Guess HashSet | O(n) | O(n) | Optimal approach for interviews and large inputs |
| Union-Find Assisted Validation | O(n α(n)) | O(n) | When combining connectivity checks with structural constraints |
Frequently asked questions
Is Count Number of Possible Root Nodes easy or hard?
LeetCode classifies this problem as Hard because it requires a rerooting dynamic programming technique on trees. The core difficulty is realizing that changing the root only affects the orientation of a single edge and that the score can be updated incrementally.
Count Number of Possible Root Nodes Python/Java solution
Implement the optimal approach using adjacency lists and a hash set for guesses. Run two DFS traversals: the first counts correct guesses from an initial root, and the second reroots the tree while adjusting the score. The algorithm works the same in Python, Java, C++, and JavaScript with O(n) complexity.
How to solve Count Number of Possible Root Nodes in O(n)?
Store all guesses in a hash set and build an adjacency list for the tree. Run a DFS from node 0 to count correct parent-child guesses. Then perform a second DFS that reroots the tree: when moving the root from u to v, subtract the guess (u, v) if present and add (v, u) if present. This constant-time adjustment per edge yields an O(n) solution.
What is the best approach for Count Number of Possible Root Nodes?
The most efficient method uses DFS with a rerooting technique. First compute the number of correct guesses assuming node 0 is the root. Then reroot the tree across each edge and update the count by adjusting only the affected guess directions. This runs in O(n) time with O(n) space.
Is Count Number of Possible Root Nodes asked at Google/Amazon/Meta?
Tree rerooting and DFS-based dynamic programming problems appear frequently in interviews at companies like Google, Amazon, and Meta. This problem specifically tests graph traversal, hash-based lookups, and rerooting DP techniques that are common in senior-level coding interviews.
What data structure is used in Count Number of Possible Root Nodes?
The solution relies on an adjacency list to represent the tree and a hash set to store guesses for O(1) lookups. Depth-first search is used for traversal, and reroot dynamic programming updates the correct guess count while moving the root across edges.
What is the time complexity of Count Number of Possible Root Nodes?
The optimal solution runs in O(n) time where n is the number of nodes. Two DFS traversals are used: one to count correct guesses for an initial root and another to reroot and update counts incrementally. Space complexity is O(n) for adjacency lists and the guess hash set.