There is a tree (i.e., a connected, undirected graph that has no cycles) consisting of n nodes numbered from 0 to n - 1 and exactly n - 1 edges. Each node has a value associated with it, and the root of the tree is node 0.

To represent this tree, you are given an integer array nums and a 2D array edges. Each nums[i] represents the ith node’s value, and each edges[j] = [uj, vj] represents an edge between nodes uj and vj in the tree.

Two values x and y are coprime if gcd(x, y) == 1 where gcd(x, y) is the greatest common divisor of x and y.

An ancestor of a node i is any other node on the shortest path from node i to the root. A node is not considered an ancestor of itself.

Return an array ans of size nwhere ans[i] is the closest ancestor to node i such that nums[i] and nums[ans[i]] are coprime, or -1 if there is no such ancestor.

Example 1:

Input: nums = [2,3,3,2], edges = [[0,1],[1,2],[1,3]]
Output: [-1,0,0,1]
Explanation: In the above figure, each node's value is in parentheses.
- Node 0 has no coprime ancestors.
- Node 1 has only one ancestor, node 0. Their values are coprime (gcd(2,3) == 1).
- Node 2 has two ancestors, nodes 1 and 0. Node 1's value is not coprime (gcd(3,3) == 3), but node 0's
value is (gcd(2,3) == 1), so node 0 is the closest valid ancestor.
- Node 3 has two ancestors, nodes 1 and 0. It is coprime with node 1 (gcd(3,2) == 1), so node 1 is its
closest valid ancestor.

Example 2:

Input: nums = [5,6,10,2,3,6,15], edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]]
Output: [-1,0,-1,0,0,0,-1]

Constraints:

• nums.length == n
• 1 <= nums[i] <= 50
• 1 <= n <= 105
• edges.length == n - 1
• edges[j].length == 2
• 0 <= uj, vj < n
• uj != vj

## Efficient Solutions by OpHaxor:

public int[] getCoprimes(int[] nums, int[][] edges) {
int n = nums.length;
for (int[] e : edges) {
}
ans = new int[n];
for (int i = 1; i <= 50; i++) {
for (int j = 1; j <= 50; j++) {
coprime[i][j] = gcd(i, j) == 1;
}
}
stacks = new ArrayList<>(51);
for (int i = 0; i <= 50; i++) stacks.add(new ArrayDeque<int[]>());
return ans;
}

int[] ans;
boolean[][] coprime = new boolean[51][51];
// id, depth
List<ArrayDeque<int[]>> stacks;

void dfs(int[] nums, List<List<Integer>> adj, int u, int p, int d) {
int me = nums[u];
int[] best = new int[]{-1, -1};
for (int other = 1; other <= 50; other++) {
if (!coprime[me][other]) continue;
if (!stacks.get(other).isEmpty()) {
int[] anc = stacks.get(other).peekLast();
if (anc[1] > best[1]) {
best[0] = anc[0];
best[1] = anc[1];
}
}
}
ans[u] = best[0];
for (int v : adj.get(u)) {
if (v == p) continue;
}
stacks.get(me).removeLast();
}

int gcd(int a, int b) {
return a == 0 ? b : gcd(b % a, a);
}
def getCoprimes(self, nums: List[int], edges: List[List[int]]) -> List[int]:
length = len(nums)

graph = [[] for i in range(length)]
for first, second in edges:
graph[first].append(second)
graph[second].append(first)

answers = [-1 for _ in range(length)]
path = defaultdict(list)

def dfs(vertex: int = 0, parent: int = -1, depth: int = 0):
best_depth = -1
for value, parents in path.items():
if math.gcd(value, nums[vertex]) == 1 and parents:
if parents[-1][1] > best_depth:
best_depth = parents[-1][1]

path[nums[vertex]].append((vertex, depth))

path[nums[vertex]].pop()

dfs()

vector<int> getCoprimes(vector<int>& nums, vector<vector<int>>& edges) {
int n = nums.size();
vector<vector<int>> neighbours(n);
for(const auto& edge : edges){
neighbours[edge[0]].push_back(edge[1]);
neighbours[edge[1]].push_back(edge[0]);
}
vector<int> level(n, 0);
// level order traversal
queue<int> q;
q.push(0);
vector<bool> visited(n, false);
visited[0] = true;
int depth = 0;
while(!q.empty()){
int size = q.size();
while(size--){
int cur = q.front();
q.pop();
level[cur] = depth;
for(int next : neighbours[cur]){
if(!visited[next]){
visited[next] = true;
q.push(next);
}
}
}
depth++;
}

vector<int> ans(n, -1);
visited = vector<bool>(n, false);
vector<stack<int>> lastSeen(51);
dfs(neighbours, nums, level, 0, visited, lastSeen, ans);
return ans;
}
private:
int gcd(int a, int b){
if(a == 0){
return b;
}
if(b == 0){
return a;
}
return gcd(b, a%b);
}

void dfs(const vector<vector<int>>& neighbours, const vector<int>& nums, const vector<int>& level, int cur, vector<bool>& visited, vector<stack<int>>& lastSeen, vector<int>& ans){
for(int i = 1; i < lastSeen.size(); i++){
if(!lastSeen[i].empty()){
int j = lastSeen[i].top();
if(gcd(nums[cur], nums[j]) == 1){
if(ans[cur] == -1 || level[ans[cur]] < level[j]){
ans[cur] = j;
}
}
}
}
lastSeen[nums[cur]].push(cur);
visited[cur] = true;
for(int next : neighbours[cur]){
if(!visited[next]){
dfs(neighbours, nums, level, next, visited, lastSeen, ans);
}
}
lastSeen[nums[cur]].pop();
}
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