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.
Since the range of $nums[i]$ in the problem is $[1, 50]$, we can preprocess all the coprime numbers for each number and record them in the array $f$, where $f[i]$ represents all the coprime numbers of $i$.
Next, we can use a backtracking method to traverse the entire tree from the root node. For each node $i$, we can get all the coprime numbers of $nums[i]$ through the array $f$. Then we enumerate all the coprime numbers of $nums[i]$, find the ancestor node $t$ that has appeared and has the maximum depth, which is the nearest coprime ancestor node of $i$. Here we can use a stack array $stks$ of length $51$ to get each appeared value $v$ and its depth. The top element of each stack $stks[v]$ is the nearest ancestor node with the maximum depth.
The time complexity is $O(n \times M)$, and the space complexity is $O(M^2 + n)$. Where $n$ is the number of nodes, and $M$ is the maximum value of $nums[i]$, in this problem $M = 50$.