Maximum Number of K-Divisible Components

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Ah, the “Maximum Number of K-Divisible Components” problem! It’s like trying to find the perfect number of slices in a pizza so that everyone gets a piece without any leftovers. You know, the kind of math that makes you question your life choices while you’re just trying to enjoy a good meal.

Imagine you’re at a party, and you have a group of friends who are all on different diets. Some are counting calories, others are just trying to avoid gluten, and a few are on a strict “no fun” diet. Now, you have to figure out how to divide the pizza in such a way that everyone is satisfied, and you don’t end up with a slice that’s just too much for anyone to handle. That’s pretty much what this problem is about!

In this problem, you are given a tree (yes, not the kind you climb, but a data structure) and you need to find the maximum number of connected components whose sum of values is divisible by a given integer k. Sounds simple, right? Well, it’s a bit more complicated than it seems, but don’t worry, we’ve got the code to help you out!

Java Solution


class Solution {
  public int maxKDivisibleComponents(int n, int[][] edges, int[] values, int k) {
    List[] graph = new List[n];

    for (int i = 0; i < n; i++)
      graph[i] = new ArrayList<>();

    for (int[] edge : edges) {
      final int u = edge[0];
      final int v = edge[1];
      graph[u].add(v);
      graph[v].add(u);
    }

    dfs(graph, 0, /*prev=*/-1, values, k);
    return ans;
  }

  private int ans = 0;

  private long dfs(List[] graph, int u, int prev, int[] values, int k) {
    long treeSum = values[u];

    for (int v : graph[u])
      if (v != prev)
        treeSum += dfs(graph, v, u, values, k);

    if (treeSum % k == 0)
      ++ans;
    return treeSum;
  }
}

Approach Explanation

The code uses Depth-First Search (DFS) to traverse the tree structure. Starting from the root node (node 0), it calculates the sum of values for each subtree. If the sum of any subtree is divisible by k, it increments the count of valid components. This way, it efficiently checks all possible components without needing to explicitly create them.

Time and Space Complexity

Complexity Type	Complexity
Time Complexity	O(n) – Each node is visited once.
Space Complexity	O(n) – The space used for the graph representation and the recursion stack.

Real-World Example

Let’s say you’re organizing a charity event and you have a group of volunteers. Each volunteer has a different skill set (represented by values), and you want to form teams (components) such that the total skill level of each team is divisible by a certain number k. This way, you can ensure that each team is balanced and can tackle tasks effectively. The problem is essentially about finding the maximum number of such balanced teams.