Formatted question description: https://leetcode.ca/all/1766.html

1766. Tree of Coprimes

Level

Hard

Description

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 i-th node’s value, and each edges[j] = [u_j, v_j] represents an edge between nodes u_j and v_j 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:

Image text

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:

Image text

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 <= u_j, v_j < n
  • u_j != v_j

Solution

First, store each node’s adjacent nodes. Next, preprocess the numbers from 1 to 50 and find coprimes for each number. Then, do depth first search starting from root node 0 and fimd the nearest coprime node for each node.

class Solution {
    int[] ans;
    Map<Integer, List<Integer>> edgesMap = new HashMap<Integer, List<Integer>>();
    Map<Integer, List<Integer>> coprimesMap = new HashMap<Integer, List<Integer>>();
    int[] depths;
    int[] pos = new int[51];

    public int[] getCoprimes(int[] nums, int[][] edges) {
        int n = nums.length;
        ans = new int[n];
        depths = new int[n];
        Arrays.fill(ans, -1);
        Arrays.fill(pos, -1);
        for (int[] edge : edges) {
            int node0 = edge[0], node1 = edge[1];
            List<Integer> list0 = edgesMap.getOrDefault(node0, new ArrayList<Integer>());
            List<Integer> list1 = edgesMap.getOrDefault(node1, new ArrayList<Integer>());
            list0.add(node1);
            list1.add(node0);
            edgesMap.put(node0, list0);
            edgesMap.put(node1, list1);
        }
        for (int i = 1; i <= 50; i++) {
            for (int j = 1; j <= 50; j++) {
                if (gcd(i, j) == 1) {
                    List<Integer> list = coprimesMap.getOrDefault(i, new ArrayList<Integer>());
                    list.add(j);
                    coprimesMap.put(i, list);
                }
            }
        }
        depthFirstSearch(nums, 0, -1);
        return ans;
    }

    public void depthFirstSearch(int[] nums, int u, int form) {
        int t = nums[u];
        for (int v : coprimesMap.get(t)) {
            if (pos[v] != -1) {
                if (ans[u] == -1 || depths[ans[u]] < depths[pos[v]])
                    ans[u] = pos[v];
            }
        }
        int p = pos[t];
        pos[t] = u;
        for (int i : edgesMap.get(u)) {
            if (i != form) {
                depths[i] = depths[u] + 1;
                depthFirstSearch(nums, i, u);
            }
        }
        pos[t] = p;
    }

    public int gcd(int a, int b) {
        while (a != 0 && b != 0) {
            if (a >= b)
                a %= b;
            else
                b %= a;
        }
        return a == 0 ? b : a;
    }
}

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