1766. Tree of Coprimes

Problem

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 **n, **where 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 <= 10^5

• edges.length == n - 1

• edges[j].length == 2

• 0 <= uj, vj < n

• uj != vj

Solution

class Solution {
    private void dfs(
            int[] v2n,
            int[] v2d,
            int depth,
            int parent,
            int node,
            int[] ans,
            int[] nums,
            ArrayList<Integer>[] neighbors) {
        int d = Integer.MIN_VALUE;
        int n = -1;
        int v = nums[node];
        for (int i = 1; i <= 50; i++) {
            if (v2n[i] != -1 && v2d[i] > d && gcd(i, v) == 1) {
                d = v2d[i];
                n = v2n[i];
            }
        }
        ans[node] = n;
        int v2NOld = v2n[v];
        int v2DOld = v2d[v];
        v2n[v] = node;
        v2d[v] = depth;
        for (int child : neighbors[node]) {
            if (child == parent) {
                continue;
            }
            dfs(v2n, v2d, depth + 1, node, child, ans, nums, neighbors);
        }
        v2n[v] = v2NOld;
        v2d[v] = v2DOld;
    }

    private int gcd(int x, int y) {
        return x == 0 ? y : gcd(y % x, x);
    }

    public int[] getCoprimes(int[] nums, int[][] edges) {
        int n = nums.length;
        ArrayList<Integer>[] neighbors = new ArrayList[n];
        for (int i = 0; i < n; i++) {
            neighbors[i] = new ArrayList<>();
        }
        for (int[] edge : edges) {
            neighbors[edge[0]].add(edge[1]);
            neighbors[edge[1]].add(edge[0]);
        }
        int[] ans = new int[n];
        int[] v2n = new int[51];
        int[] v2d = new int[51];
        Arrays.fill(v2n, -1);
        dfs(v2n, v2d, 0, -1, 0, ans, nums, neighbors);
        return ans;
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).