Introduction

In this blog post, we will explore how to solve the LeetCode problem "1766" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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

Explanation

Here's the solution to the problem, incorporating the requested elements:

• High-Level Approach:

  • Perform a Depth-First Search (DFS) to traverse the tree.
  • Maintain a stack (or similar data structure) to track ancestors at each node during the DFS. This stack stores node values and their depths.
  • For each node, iterate through the ancestor stack, check for coprimality, and find the closest coprime ancestor.

• Complexity:

  • Time Complexity: O(n * log(M)), where n is the number of nodes and M is the maximum value of nums[i] (in this case, 50). The log(M) factor comes from the gcd calculation.
  • Space Complexity: O(n), mainly for the adjacency list, the ans array, and the ancestor stack in the DFS.

Code

    from math import gcd

def get_coprime_ancestors(nums, edges):
    n = len(nums)
    adj = [[] for _ in range(n)]
    for u, v in edges:
        adj[u].append(v)
        adj[v].append(u)

    ans = [-1] * n
    ancestor_stack = []  # (node, depth)

    def dfs(node, parent, depth):
        closest_ancestor = -1
        min_depth = float('inf')

        for i in range(len(ancestor_stack) - 1, -1, -1):
            ancestor, ancestor_depth = ancestor_stack[i]
            if gcd(nums[node], nums[ancestor]) == 1:
                if ancestor_depth < min_depth:
                    min_depth = ancestor_depth
                    closest_ancestor = ancestor

        ans[node] = closest_ancestor

        ancestor_stack.append((node, depth))
        for neighbor in adj[node]:
            if neighbor != parent:
                dfs(neighbor, node, depth + 1)
        ancestor_stack.pop()  # Backtrack

    dfs(0, -1, 0)
    return ans