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(nΓM), and the space complexity is O(M2+n). Where n is the number of nodes, and M is the maximum value of nums[i], in this problem M=50.