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Maximum Number of K-Divisible Components

Difficulty: Hard

Problem Description

You are given an undirected tree with n nodes, represented by edges, and an array of values associated with each node. You need to determine the maximum number of components that can be formed by removing edges from the tree such that the sum of the values in each component is divisible by a given integer k.

Key Insights

  • A valid split can be achieved by removing edges, resulting in connected components.
  • The value of a component is the sum of the values of its nodes.
  • The sum of values in each component must be divisible by k.
  • Depth-First Search (DFS) can be used to traverse the tree and calculate component values.
  • A counter can be maintained to track the number of valid components formed.

Space and Time Complexity

Time Complexity: O(n), where n is the number of nodes, as we traverse each node once. Space Complexity: O(n), for storing the adjacency list representation of the tree and the recursion stack in DFS.

Solution

To solve this problem, we can use a Depth-First Search (DFS) approach:

  1. Construct an adjacency list from the edges to represent the tree.
  2. Use a DFS function to calculate the sum of values for each subtree rooted at a node.
  3. If the sum of a subtree is divisible by k, we increment our valid component counter and return 0 (indicating that this subtree can be treated as a separate component).
  4. If not, we return the sum of that subtree.
  5. The main function initializes the necessary data structures and calls the DFS function starting from the root node.

Code Solutions

def maxKDivisibleComponents(n, edges, values, k):
    from collections import defaultdict

    # Build the adjacency list for the tree
    tree = defaultdict(list)
    for a, b in edges:
        tree[a].append(b)
        tree[b].append(a)

    # Variable to count the number of valid components
    count = 0

    def dfs(node, parent):
        nonlocal count
        current_sum = values[node]

        for neighbor in tree[node]:
            if neighbor != parent:  # Avoid going back to the parent
                subtree_sum = dfs(neighbor, node)
                current_sum += subtree_sum

        # Check if the current component's sum is divisible by k
        if current_sum % k == 0:
            count += 1
            return 0  # Return 0 to signify this component can be split
        return current_sum  # Return the current sum to the parent

    dfs(0, -1)  # Start DFS from the root node (0) with no parent (-1)
    
    return count
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