How do you solve Tree of Coprimes on LeetCode?

Tree of Coprimes (LeetCode #1766) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a HashMap to track last seen values can significantly reduce the number of checks needed.

What is the brute force solution for Tree of Coprimes?

The brute force approach for Tree of Coprimes is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each node's ancestors one by one to find the closest coprime ancestor. This is straightforward but inefficient, especially for larger trees. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Tree of Coprimes?

Tree of Coprimes uses the following concepts: Array, Math, Tree, Depth-First Search, Number Theory. The recommended patterns to study are: Hash Map, Depth-First Search, Number Theory.

What are the interview tips for Tree of Coprimes?

Explain your thought process clearly while coding. Consider edge cases and discuss them with the interviewer. Be prepared to optimize your solution and explain the reasoning behind your improvements.

What are common mistakes when solving Tree of Coprimes?

Not using a data structure to track last seen values, leading to inefficient ancestor checks. Failing to handle edge cases like nodes with the same value.

#1766

Tree of Coprimes

Hard

ArrayMathTreeDepth-First SearchNumber TheoryHash MapDepth-First SearchNumber Theory

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a depth-first search (DFS) while maintaining a map of the last seen node values. This allows us to efficiently find the closest coprime ancestor without redundant checks.

⚙️

Algorithm

3 steps

1Step 1: Create a graph representation of the tree using adjacency lists.
2Step 2: Use a DFS to traverse the tree, maintaining a HashMap to track the last seen node for each value.
3Step 3: For each node, check the HashMap for coprime values and update the answer accordingly.

solution.py23 lines

1from math import gcd
2from collections import defaultdict
3
4def coprime_ancestors(nums, edges):
5    n = len(nums)
6    graph = [[] for _ in range(n)]
7    for u, v in edges:
8        graph[u].append(v)
9        graph[v].append(u)
10    ans = [-1] * n
11    last_seen = defaultdict(int)
12    def dfs(node, parent):
13        for value in last_seen:
14            if gcd(nums[node], value) == 1:
15                ans[node] = last_seen[value]
16                break
17        last_seen[nums[node]] = node
18        for neighbor in graph[node]:
19            if neighbor != parent:
20                dfs(neighbor, node)
21        del last_seen[nums[node]]
22    dfs(0, -1)
23    return ans

ℹ

Complexity note: The complexity is O(n) because we visit each node once and maintain a HashMap for the last seen nodes, which allows for efficient lookups.

1Using a HashMap to track last seen values can significantly reduce the number of checks needed.
2Understanding the properties of coprime numbers and gcd is crucial for optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.