Leetcode #2872: Maximum Number of K-Divisible Components
In this guide, we solve Leetcode #2872 Maximum Number of K-Divisible Components 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 an undirected tree with n nodes labeled from 0 to n - 1. You are given the integer n and a 2D integer array edges of length n - 1, where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Tree, Depth-First Search
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: n = 5, edges = [[0,2],[1,2],[1,3],[2,4]], values = [1,8,1,4,4], k = 6
Output: 2
Explanation: We remove the edge connecting node 1 with 2. The resulting split is valid because:
- The value of the component containing nodes 1 and 3 is values[1] + values[3] = 12.
- The value of the component containing nodes 0, 2, and 4 is values[0] + values[2] + values[4] = 6.
It can be shown that no other valid split has more than 2 connected components.
Python Solution
class Solution:
def maxKDivisibleComponents(
self, n: int, edges: List[List[int]], values: List[int], k: int
) -> int:
def dfs(i: int, fa: int) -> int:
s = values[i]
for j in g[i]:
if j != fa:
s += dfs(j, i)
nonlocal ans
ans += s % k == 0
return s
g = [[] for _ in range(n)]
for a, b in edges:
g[a].append(b)
g[b].append(a)
ans = 0
dfs(0, -1)
return ans
Complexity
The time complexity is and the space complexity is , where is the number of nodes in the tree. The space complexity is , where is the number of nodes in the tree.
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.