Leetcode #1766: Tree of Coprimes
In this guide, we solve Leetcode #1766 Tree of Coprimes in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Tree, Depth-First Search, Array, Math, Number Theory
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
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.
Python Solution
class Solution:
def getCoprimes(self, nums: List[int], edges: List[List[int]]) -> List[int]:
def dfs(i, fa, depth):
t = k = -1
for v in f[nums[i]]:
stk = stks[v]
if stk and stk[-1][1] > k:
t, k = stk[-1]
ans[i] = t
for j in g[i]:
if j != fa:
stks[nums[i]].append((i, depth))
dfs(j, i, depth + 1)
stks[nums[i]].pop()
g = defaultdict(list)
for u, v in edges:
g[u].append(v)
g[v].append(u)
f = defaultdict(list)
for i in range(1, 51):
for j in range(1, 51):
if gcd(i, j) == 1:
f[i].append(j)
stks = defaultdict(list)
ans = [-1] * len(nums)
dfs(0, -1, 0)
return ans
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.