LeetCode 2872 - Maximum Number of K-Divisible Components

This problem is asking us to split a given undirected tree into as many connected components as possible, under the condition that the sum of values of nodes in each component is divisible by a given integer k.

leetcodehardtreedepth-first-search

LeetCode Problem 2872

Difficulty: 🔴 Hard
Topics: Tree, Depth-First Search

Solution

Problem Understanding

This problem is asking us to split a given undirected tree into as many connected components as possible, under the condition that the sum of values of nodes in each component is divisible by a given integer k. The tree is represented with n nodes numbered from 0 to n - 1 and a list of edges, where each edge connects two nodes. Each node has an associated value given in the values array. A "valid split" is defined by cutting edges so that every resulting connected component's node values sum to a multiple of k. The output should be the maximum number of components achievable under this constraint.

The input constraints ensure that:

  1. The tree has up to 3 * 10^4 nodes, so brute-force approaches exploring all edge-removal combinations are infeasible.
  2. Each node's value is non-negative and k is at least 1.
  3. The sum of all node values is divisible by k, guaranteeing that at least one valid split exists, namely the whole tree itself.

Important edge cases include trees where no splits are possible beyond the whole tree, trees where every node's value is divisible by k, or very unbalanced trees that might affect recursion depth in DFS implementations.

Approaches

Brute Force

A brute-force solution would attempt to generate all possible ways to remove edges from the tree. For each configuration, we would compute the sum of values for every connected component and check if all sums are divisible by k. While this would yield the correct result, it is completely infeasible given the constraints, because there are 2^(n-1) possible edge removal combinations.

Optimal Insight

The key insight is that we do not need to try every possible removal. Instead, we can use Depth-First Search (DFS) and compute the sum of values for each subtree. If the sum of a subtree is divisible by k, we can safely remove the edge connecting this subtree to its parent because the subtree forms a valid component on its own. By recursively applying this logic, we can count the maximum number of valid components in a single DFS traversal.

The process can be formalized as:

  1. Traverse the tree in DFS from any node (usually 0 as root).
  2. At each node, compute the sum of the subtree rooted at that node.
  3. If the subtree sum is divisible by k and it is not the root, increment a counter representing valid components and do not propagate the sum further to its parent (since it is now an independent component).
  4. Return the sum for non-divisible subtrees so they contribute to their parent's sum.

This approach guarantees a single pass over the tree and achieves linear time complexity.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Try all combinations of edge removals; infeasible for large n
Optimal O(n) O(n) DFS computes subtree sums and counts divisible components

Algorithm Walkthrough

  1. Represent the tree using an adjacency list for efficient traversal.
  2. Initialize a counter components to zero, which will track the number of valid components found.
  3. Define a recursive DFS function that takes the current node and its parent as parameters.
  4. Within DFS, start with the current node's value as the initial subtree sum.
  5. Iterate over all child nodes (neighbors not equal to the parent) and recursively compute their subtree sums.
  6. Add each child subtree sum to the current node's sum.
  7. After processing all children, check if the subtree sum is divisible by k. If yes, and the node is not the root, increment components and return 0 to the parent (the component is now independent). Otherwise, return the subtree sum.
  8. Start DFS from node 0 with parent -1.
  9. Return the final count of components.

Why it works: The invariant is that whenever a subtree sum is divisible by k, it can form a valid independent component. By recursively applying this from leaves to root, we maximize the number of divisible components without double-counting or violating divisibility rules.

Python Solution

from typing import List
from collections import defaultdict

class Solution:
    def maxKDivisibleComponents(self, n: int, edges: List[List[int]], values: List[int], k: int) -> int:
        tree = defaultdict(list)
        for u, v in edges:
            tree[u].append(v)
            tree[v].append(u)
        
        self.components = 0
        
        def dfs(node: int, parent: int) -> int:
            subtotal = values[node]
            for neighbor in tree[node]:
                if neighbor == parent:
                    continue
                subtotal += dfs(neighbor, node)
            if subtotal % k == 0 and parent != -1:
                self.components += 1
                return 0
            return subtotal
        
        dfs(0, -1)
        return self.components

Implementation walkthrough: The adjacency list tree stores all neighbors. The recursive dfs computes subtree sums. If a subtree is divisible by k and is not the root, it increments self.components and returns zero to detach it from the parent. Finally, self.components gives the maximum number of K-divisible components.

Go Solution

func maxKDivisibleComponents(n int, edges [][]int, values []int, k int) int {
    tree := make([][]int, n)
    for _, e := range edges {
        u, v := e[0], e[1]
        tree[u] = append(tree[u], v)
        tree[v] = append(tree[v], u)
    }
    
    components := 0
    
    var dfs func(node, parent int) int
    dfs = func(node, parent int) int {
        subtotal := values[node]
        for _, neighbor := range tree[node] {
            if neighbor == parent {
                continue
            }
            subtotal += dfs(neighbor, node)
        }
        if subtotal % k == 0 && parent != -1 {
            components++
            return 0
        }
        return subtotal
    }
    
    dfs(0, -1)
    return components
}

Go-specific notes: The tree is represented as a slice of slices. DFS is defined as a closure to capture components by reference. Nil slices are handled automatically by Go during append. Integer arithmetic is safe due to constraints.

Worked Examples

Example 1: n = 5, edges = [[0,2],[1,2],[1,3],[2,4]], values = [1,8,1,4,4], k = 6

DFS order and subtree sums:

Node Subtree sum Action
3 4 not divisible by 6, return 4
1 8 + 4 = 12 divisible by 6, increment components = 1, return 0
4 4 return 4
2 1 + 0 + 4 = 5 not divisible, return 5
0 1 + 5 = 6 divisible by 6, root node, do not increment, return 6

Result: components = 2.

Example 2: n = 7, edges = [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]], values = [3,0,6,1,5,2,1], k = 3

DFS computations:

Node Subtree sum Action
3 1 return 1
4 5 return 5
1 0 + 1 + 5 = 6 divisible, increment components = 1, return 0
5 2 return 2
6 1 return 1
2 6 + 2 + 1 = 9 divisible, increment components = 2, return 0
0 3 + 0 + 0 = 3 divisible, root, do not increment

Result: components = 3.

Complexity Analysis

Measure Complexity Explanation
Time O(n) Each node is visited once in DFS
Space O(n) Adjacency list and recursion stack use O(n)

The linear time complexity is optimal for this problem given the tree size. The recursion stack is bounded by the height of the tree, which in the worst case is O(n).

Test Cases

# Provided examples
assert Solution().maxKDivisibleComponents(5, [[0,2],[1,2],[1,3],[2,4]], [1,8,1,4,4], 6) == 2
assert Solution().maxKDivisibleComponents(7, [[0,1],[0,2],[1,3],[1,4],[2,5],[2,6]], [3,0,6,1,5,2,1], 3) == 3

# Edge cases
assert Solution().maxKDivisibleComponents(1, [], [5], 5) == 0  # Single node tree
assert Solution().maxKDivisibleComponents(3, [[0,1],[1,2]], [2,2,2], 2) == 2