# 3004. Maximum Subtree of the Same Color

## Description

You are given a 2D integer array edges representing a tree with n nodes, numbered from 0 to n - 1, rooted at node 0, where edges[i] = [ui, vi] means there is an edge between the nodes vi and ui.

You are also given a 0-indexed integer array colors of size n, where colors[i] is the color assigned to node i.

We want to find a node v such that every node in the subtree of v has the same color.

Return the size of such subtree with the maximum number of nodes possible.

Example 1:

Input: edges = [[0,1],[0,2],[0,3]], colors = [1,1,2,3]
Output: 1
Explanation: Each color is represented as: 1 -> Red, 2 -> Green, 3 -> Blue. We can see that the subtree rooted at node 0 has children with different colors. Any other subtree is of the same color and has a size of 1. Hence, we return 1.


Example 2:

Input: edges = [[0,1],[0,2],[0,3]], colors = [1,1,1,1]
Output: 4
Explanation: The whole tree has the same color, and the subtree rooted at node 0 has the most number of nodes which is 4. Hence, we return 4.


Example 3:

Input: edges = [[0,1],[0,2],[2,3],[2,4]], colors = [1,2,3,3,3]
Output: 3
Explanation: Each color is represented as: 1 -> Red, 2 -> Green, 3 -> Blue. We can see that the subtree rooted at node 0 has children with different colors. Any other subtree is of the same color, but the subtree rooted at node 2 has a size of 3 which is the maximum. Hence, we return 3.


Constraints:

• n == edges.length + 1
• 1 <= n <= 5 * 104
• edges[i] == [ui, vi]
• 0 <= ui, vi < n
• colors.length == n
• 1 <= colors[i] <= 105
• The input is generated such that the graph represented by edges is a tree.

## Solutions

### Solution 1: DFS

First, according to the edge information given in the problem, we construct an adjacency list $g$, where $g[a]$ represents all adjacent nodes of node $a$. Then we create an array $size$ of length $n$, where $size[a]$ represents the number of nodes in the subtree with node $a$ as the root.

Next, we design a function $dfs(a, fa)$, which will return whether the subtree with node $a$ as the root meets the requirements of the problem. The execution process of the function $dfs(a, fa)$ is as follows:

• First, we use a variable $ok$ to record whether the subtree with node $a$ as the root meets the requirements of the problem, and initially $ok$ is $true$.
• Then, we traverse all adjacent nodes $b$ of node $a$. If $b$ is not the parent node $fa$ of $a$, then we recursively call $dfs(b, a)$, save the return value into the variable $t$, and update $ok$ to the value of $ok$ and $colors[a] = colors[b] \land t$, where $\land$ represents logical AND operation. Then, we update $size[a] = size[a] + size[b]$.
• After that, we judge the value of $ok$. If $ok$ is $true$, then we update the answer $ans = \max(ans, size[a])$.
• Finally, we return the value of $ok$.

We call $dfs(0, -1)$, where $0$ represents the root node number, and $-1$ represents that the root node has no parent node. The final answer is $ans$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Where $n$ is the number of nodes.

• class Solution {
private List<Integer>[] g;
private int[] colors;
private int[] size;
private int ans;

public int maximumSubtreeSize(int[][] edges, int[] colors) {
int n = edges.length + 1;
g = new List[n];
size = new int[n];
this.colors = colors;
Arrays.fill(size, 1);
Arrays.setAll(g, i -> new ArrayList<>());
for (var e : edges) {
int a = e[0], b = e[1];
}
dfs(0, -1);
return ans;
}

private boolean dfs(int a, int fa) {
boolean ok = true;
for (int b : g[a]) {
if (b != fa) {
boolean t = dfs(b, a);
ok = ok && colors[a] == colors[b] && t;
size[a] += size[b];
}
}
if (ok) {
ans = Math.max(ans, size[a]);
}
return ok;
}
}

• class Solution {
public:
int maximumSubtreeSize(vector<vector<int>>& edges, vector<int>& colors) {
int n = edges.size() + 1;
vector<int> g[n];
vector<int> size(n, 1);
for (auto& e : edges) {
int a = e[0], b = e[1];
g[a].push_back(b);
g[b].push_back(a);
}
int ans = 0;
function<bool(int, int)> dfs = [&](int a, int fa) {
bool ok = true;
for (int b : g[a]) {
if (b != fa) {
bool t = dfs(b, a);
ok = ok && colors[a] == colors[b] && t;
size[a] += size[b];
}
}
if (ok) {
ans = max(ans, size[a]);
}
return ok;
};
dfs(0, -1);
return ans;
}
};

• class Solution:
def maximumSubtreeSize(self, edges: List[List[int]], colors: List[int]) -> int:
def dfs(a: int, fa: int) -> bool:
ok = True
for b in g[a]:
if b != fa:
t = dfs(b, a)
ok = ok and colors[a] == colors[b] and t
size[a] += size[b]
if ok:
nonlocal ans
ans = max(ans, size[a])
return ok

n = len(edges) + 1
g = [[] for _ in range(n)]
size = [1] * n
for a, b in edges:
g[a].append(b)
g[b].append(a)
ans = 0
dfs(0, -1)
return ans


• func maximumSubtreeSize(edges [][]int, colors []int) (ans int) {
n := len(edges) + 1
g := make([][]int, n)
for _, e := range edges {
a, b := e[0], e[1]
g[a] = append(g[a], b)
g[b] = append(g[b], a)
}
size := make([]int, n)
var dfs func(int, int) bool
dfs = func(a, fa int) bool {
size[a] = 1
ok := true
for _, b := range g[a] {
if b != fa {
t := dfs(b, a)
ok = ok && t && colors[a] == colors[b]
size[a] += size[b]
}
}
if ok {
ans = max(ans, size[a])
}
return ok
}
dfs(0, -1)
return
}

• function maximumSubtreeSize(edges: number[][], colors: number[]): number {
const n = edges.length + 1;
const g: number[][] = Array.from({ length: n }, () => []);
for (const [a, b] of edges) {
g[a].push(b);
g[b].push(a);
}
const size: number[] = Array(n).fill(1);
let ans = 0;
const dfs = (a: number, fa: number): boolean => {
let ok = true;
for (const b of g[a]) {
if (b !== fa) {
const t = dfs(b, a);
ok = ok && t && colors[a] === colors[b];
size[a] += size[b];
}
}
if (ok) {
ans = Math.max(ans, size[a]);
}
return ok;
};
dfs(0, -1);
return ans;
}