LeetCode 2872: Maximum Number of K-Divisible Components Solution

Master LeetCode problem 2872 (Maximum Number of K-Divisible Components), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2872. Maximum Number of K-Divisible Components

Hard

Tree Depth-First Search

Problem Explanation

Explanation

To solve this problem, we can perform a depth-first search (DFS) on the tree and keep track of the total sum of values in each subtree. At each node, we can calculate the sum of values in the subtree rooted at that node. If the sum is divisible by k, we can split the subtree at that node.

We will recursively visit each node in the tree, update the sum of values at each node, and count the number of connected components that satisfy the condition. We will also keep track of the maximum number of components found so far.

To avoid double counting, we will also maintain a parent node for each node to ensure that the same subtree is not visited multiple times.

The time complexity of this solution is O(n) where n is the number of nodes in the tree, as we perform a single DFS traversal. The space complexity is also O(n) to store the parent node, sum of values, and other information for each node.

Solution Code

class Solution {
    public int maxNumOfSubstrings(int n, int[][] edges, int[] values, int k) {
        List<List<Integer>> graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }
        for (int[] edge : edges) {
            graph.get(edge[0]).add(edge[1]);
            graph.get(edge[1]).add(edge[0]);
        }
        
        int[] parent = new int[n];
        Arrays.fill(parent, -1);
        int[] subtreeSum = new int[n];
        
        int totalSum = dfs(0, graph, values, parent, subtreeSum, k);
        
        return Math.max(1, totalSum % k == 0 ? totalSum / k : 0);
    }
    
    private int dfs(int node, List<List<Integer>> graph, int[] values, int[] parent, int[] subtreeSum, int k) {
        subtreeSum[node] = values[node];
        int components = 0;
        for (int child : graph.get(node)) {
            if (child != parent[node]) {
                parent[child] = node;
                subtreeSum[node] += dfs(child, graph, values, parent, subtreeSum, k);
            }
        }
        if (subtreeSum[node] % k == 0) {
            components++;
        }
        return subtreeSum[node];
    }
}

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