2872. Maximum Number of K-Divisible Components
Problem Description
The given problem defines an undirected tree consisting of n
nodes with each node labeled from 0
to n - 1
. The structure of the tree is described using a list of edges, where each edge is represented by a pair [a_i, b_i]
connecting the nodes a_i
and b_i
. Along with the structure, each node has an associated value provided in the array values
.
The task is to find a way to split this tree into multiple components by possibly removing some edges. The condition for a valid split is that the total value of each resulting component (sum of all node values in the component) must be divisible by a given integer k
. The ultimate goal is to maximize the number of such valid components after the split.
To state it simply, you're asked to figure out the maximum number of groups you can create from the tree's nodes where the sum of values within each group is a multiple of 'k'. We must decide which edges to remove, if any, to achieve this while keeping the most groups.
Flowchart Walkthrough
Let's analyze Leetcode 2872. Maximum Number of K-Divisible Components using the Flowchart. Here's a step-by-step breakdown:
-
Is it a graph?
- Yes: The problem likely involves a graph where edges or connections depend on numbers being divisible by K.
-
Is it a tree?
- No: The data structure might have connections that form cycles or multiple roots, which makes it not strictly a tree.
-
Is the problem related to directed acyclic graphs (DAGs)?
- No: The problem does not specify that there is any directional or acyclic constraint in forming components.
-
Is the problem related to shortest paths?
- No: The goal is to maximize the count of distinct components divisible by K, not finding shortest paths.
-
Does the problem involve connectivity?
- Yes: The primary concern here is to establish connectivity between nodes based on the divisibility by K, forming components.
-
Is the graph weighted?
- No: The connections (edges) are not based on any weights but on the divisibility condition.
Since the problem involves exploring all possibilities of forming components based on a specific condition (K-divisibility) without the complexity of weights or directed edges, DFS is usually more intuitive and manageable for this type of exhaustive searching and marking off visited nodes or groups. Depth-First Search (DFS) can be leveraged for effectively marking connected components, which is pivotal in counting the maximum number of K-divisible components.
Conclusion: The flowchart suggests using Depth-First Search for this unweighted connectivity problem to identify and count components.
Intuition
The intuitive solution approach stems from two insights:
-
Divisibility Propagation: Since the entire tree's value is divisible by
k
, breaking off any subtree whose value is also divisible byk
will leave the remaining tree with a sum still divisible byk
. This holds true recursively for any subtrees split off from the main tree. Thus, we can look for divisible subtrees to count as separate components. -
DFS Utilization: To identify subtrees conforming to the divisibility rule, we need to examine the sums of each possible subtree in the tree. This is a classic use-case for Depth-First Search (DFS). Starting from the root node (or any node, as it's a connected tree), DFS can help us explore each branch to its leaves, summing the values of nodes as we backtrack up the tree. When we find a subtree sum divisible by
k
, we mark it as a potential split point and continue searching for more.
By employing DFS, we iterate over the tree nodes while keeping a running sum of the subtree rooted at each node. Whenever the sum of a subtree equals a multiple of k
, we have found a valid component. By effectively adding these components to our count and continuing the DFS, we can determine the maximum number of valid components we can split the tree into, hence arriving at the solution to our problem.
Learn more about Tree, Depth-First Search and Dynamic Programming patterns.
Solution Approach
To implement the solution, we utilize a depth-first search (DFS) algorithm, which is well-suited for exploring tree structures and their subtrees. Here's a step-by-step breakdown of how the DFS-based algorithm is applied to achieve our goal:
-
Build the Graph Representation: Before we begin the DFS, we need to convert the edge list into a graph representation that makes it easier to traverse. In Python, this often takes the form of an adjacency list, which is a list of lists where each sublist corresponds to a node and contains the nodes that it's directly connected to. This is achieved by iterating through the
edges
array and populating the adjacency list (g
) accordingly. -
Depth-First Search (DFS) Implementation:
- A recursive
dfs
function is created, which takes a node index (i
) and its parent index (fa
) as arguments. The purpose of this function is to compute the sum of values for the subtree rooted at nodei
. - Inside the function, we begin by assigning the node's value to a variable
s
. Then, we iterate over all the direct children or neighbors of nodei
(given byg[i]
). - For each neighbor
j
, we check if it's not the parent (fa
) to avoid unnecessary backtracking, and then we recursively calldfs(j, i)
to add the sum of that subtree tos
. - After processing all children, we check if the sum
s
is divisible byk
. If it is, our globalans
counter is incremented to reflect that we've found a valid subtree (or component).
- A recursive
-
Initiate DFS and Handle Return Values:
- The DFS is initiated by calling the
dfs
function on the root node (typically node0
in a 0-indexed tree) with a parent index of-1
(since the root does not have a parent). - The
dfs
initialization ensures that we explore the entire tree, summing subtree values and finding divisible components. The DFS function naturally operates by post-order traversal, examining children before processing the current node. This is crucial since subtree sums need to be calculated from the leaves upward.
- The DFS is initiated by calling the
-
Count the Result and Minus One:
- The final count must be decremented by one because the original tree's sum is divisible by
k
and thus counts as one entire component. When the DFS includes the root's sum in the final answer, we effectively double-count this component. - Therefore, the last line of the provided solution function
maxKDivisibleComponents
returnsans - 1
to adjust for this overcounting and to provide the actual maximum number of components that can result from any valid split.
- The final count must be decremented by one because the original tree's sum is divisible by
In terms of data structures, this solution uses an adjacency list for representing the graph and relies on recursion to maintain state during the DFS traversal. As for patterns, using global variables such as ans
in conjunction with DFS is a common technique to accumulate results throughout the recursion.
The elegance of this solution lies in its efficient use of the DFS traversal to examine the entire tree and identify points where the tree can be split while simultaneously ensuring that the resultant component values meet the divisibility requirement.
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Start EvaluatorExample Walkthrough
Let's illustrate the solution approach using a small example. Suppose we have a tree consisting of 5 nodes with the following edges and values:
edges = [[0, 1], [0, 2], [1, 3], [1, 4]]
values = [4, 2, 6, 3, 3]
k = 5
The adjacency list representation would look like this based on the given edges:
g = [[1, 2], [0, 3, 4], [0], [1], [1]]
Here are the steps of applying the approach to this example:
-
First, we perform DFS starting from the root node
0
. The node values are[4, 2, 6, 3, 3]
. The total sum of all values in the tree is18
, which is not divisible byk = 5
. -
From node
0
, we DFS into its children1
and2
. Node2
is a leaf node, so it contributes its value of6
and returns it to the parent node0
. -
After exploring node
2
, we move to node1
. Node1
has children3
and4
. They are both leaf nodes and return their values3
and3
. -
Back at node
1
, we add its value2
to the sum of its subtree, which is2 + 3 + 3 = 8
. The sum8
is divisible byk = 5
, meaning we've found one valid component. We'll increment our globalans
counter to1
. -
Return to root node
0
, the subtree sums of its children have been calculated, we sum them up with node0
's own value:4 + 6 + 8 = 18
. This is again not divisible byk = 5
, so no increment toans
. -
The DFS is completed. Our global
ans
counter is1
, but we have to decrement it by one, because we're not considering the entire tree as one component. Hence, the final answer is1 - 1 = 0
. -
However, if we sum the entire tree value, it's
18
, which is still not divisible by5
. Therefore, our example doesn't allow dividing the tree into components with each sum divisible byk = 5
.
Although we couldn't split the tree into valid components in this example under the divisibility rule given by k
, the procedure demonstrates how a depth-first search helps to explore all subtrees to identify the possible splits, increment a counter when a divisible subtree is found, and understanding why we subtract one in the end.
Solution Implementation
1class Solution:
2 def maxKDivisibleComponents(self, n: int, edges: List[List[int]], values: List[int], k: int) -> int:
3 # Helper method for Depth-First Search (DFS)
4 def dfs(node: int, parent: int) -> int:
5 # Start with the value of the current node.
6 total_sum = values[node]
7
8 # Traverse through the graph.
9 for neighbor in graph[node]:
10 # If the neighbor is not the parent node, perform DFS recursively.
11 if neighbor != parent:
12 total_sum += dfs(neighbor, node)
13
14 # Increment the counter if the sum is divisible by k.
15 nonlocal num_divisible_subtrees
16 num_divisible_subtrees += total_sum % k == 0
17
18 # Return the sum for the current subtree.
19 return total_sum
20
21 # Initialize graph as adjacency list representation.
22 graph = [[] for _ in range(n)]
23 for edge in edges:
24 a, b = edge
25 graph[a].append(b)
26 graph[b].append(a)
27
28 # Counter for the number of subtrees with sums divisible by k.
29 num_divisible_subtrees = 0
30 # Perform DFS starting from node 0 assuming 0 as root node.
31 dfs(0, -1)
32
33 # Decrement to exclude the sum of the whole tree,
34 # as by the problem definition it should not be counted as a subtree.
35 return num_divisible_subtrees - (values[0] % k == 0)
36
37# Note:
38# The List type import from typing module is not included in the snippet above,
39# but it should be imported at the top of the script as follows:
40# from typing import List
41
1class Solution {
2 private int answer; // Renamed 'ans' to 'answer' for better readability
3 private List<Integer>[] graph; // Renamed 'g' to 'graph' to clarify this represents a graph
4 private int[] nodeValues; // Renamed 'values' to 'nodeValues' to avoid confusion with the method parameter 'values'
5 private int divisor; // Renamed 'k' to 'divisor' for clarity
6
7 // Method to find the maximum number of components divisible by 'k'
8 public int maxKDivisibleComponents(int n, int[][] edges, int[] values, int k) {
9 this.graph = new List[n]; // Initialize the graph as a list of integer lists
10 Arrays.setAll(this.graph, i -> new ArrayList<>()); // Create an adjacency list for each node i
11
12 // Add edges to the graph to represent the bidirectional relationships between nodes
13 for (int[] edge : edges) {
14 int nodeFrom = edge[0], nodeTo = edge[1];
15 graph[nodeFrom].add(nodeTo);
16 graph[nodeTo].add(nodeFrom);
17 }
18
19 this.nodeValues = values; // Assign the provided node values to the instance variable
20 this.divisor = k; // Store the divisor as an instance variable
21
22 dfs(0, -1); // Invoke the depth-first search from the root node (assuming node 0 as root)
23 return answer; // Return the total number of components with sum divisible by 'k'
24 }
25
26 // Depth-first search to find the components with sum of values divisible by 'k'
27 private long dfs(int nodeIndex, int parentIndex) {
28 long sum = nodeValues[nodeIndex]; // Start with the node's own value
29 // Recursively visit all the connected nodes (children) that are not the parent
30 for (int adjacentNode : graph[nodeIndex]) {
31 if (adjacentNode != parentIndex) {
32 sum += dfs(adjacentNode, nodeIndex); // Add the sum of the subtree to the current node's sum
33 }
34 }
35 // If the sum is divisible by 'k', increment the answer
36 if (sum % divisor == 0) {
37 answer++;
38 }
39 return sum; // Return the sum of the node values in the subtree rooted at this node
40 }
41}
42
1#include <vector>
2#include <functional>
3using namespace std;
4
5class Solution {
6public:
7 int maxKDivisibleComponents(int nodeCount, vector<vector<int>>& edges, vector<int>& values, int divisor) {
8 int componentCount = 0; // Used to count components with value sum divisible by k
9 vector<int> adjacencyList[nodeCount]; // Adjacency list to represent the graph
10
11 // Building the graph from edges
12 for (const auto& edge : edges) {
13 int fromNode = edge[0];
14 int toNode = edge[1];
15 adjacencyList[fromNode].push_back(toNode);
16 adjacencyList[toNode].push_back(fromNode);
17 }
18
19 // Depth First Search (DFS) lambda function to traverse the graph and calculate sums
20 function<long long(int, int)> depthFirstSearch = [&](int currentNode, int parent) {
21 long long sum = values[currentNode]; // Start with the value of the current node
22
23 // Visit all adjacent nodes except the parent
24 for (int adjacentNode : adjacencyList[currentNode]) {
25 if (adjacentNode != parent) {
26 sum += depthFirstSearch(adjacentNode, currentNode);
27 }
28 }
29
30 // Increase the count if sum of values in this component is divisible by k
31 componentCount += (sum % divisor == 0);
32 return sum; // Return the sum of values from this component
33 };
34
35 // Starting DFS from node 0 assuming 0 is always part of the graph
36 depthFirstSearch(0, -1);
37
38 // Return the count of components with value sum divisible by k
39 return componentCount;
40 }
41};
42
1function maxKDivisibleComponents(
2 nodeCount: number,
3 edges: number[][],
4 values: number[],
5 k: number,
6): number {
7 // Create an adjacency list to represent the graph
8 const graph: number[][] = Array.from({ length: nodeCount }, () => []);
9
10 // Populate the graph with edges
11 for (const [node1, node2] of edges) {
12 graph[node1].push(node2);
13 graph[node2].push(node1);
14 }
15
16 // Initialize a variable to keep track of the number of components
17 let componentCount = 0;
18
19 // Depth-First Search function to explore nodes
20 const dfs = (currentNode: number, parent: number): number => {
21 // Start with the current node's value
22 let sum = values[currentNode];
23
24 // Explore all connected nodes that are not the parent of the current node
25 for (const adjacentNode of graph[currentNode]) {
26 if (adjacentNode !== parent) {
27 // Aggregate the values from child nodes
28 sum += dfs(adjacentNode, currentNode);
29 }
30 }
31
32 // If the aggregate sum is divisible by k, increment the component count
33 if (sum % k === 0) {
34 componentCount++;
35 }
36
37 // Return the sum of values for this subtree
38 return sum;
39 };
40
41 // Begin the DFS traversal from the first node (assuming 0-indexed)
42 dfs(0, -1);
43
44 // Return the total number of components where the sum is divisible by k
45 // Note: We decrement the componentCount by 1 to exclude the sum of the entire tree
46 return componentCount - 1;
47}
48
Time and Space Complexity
The given code defines a function that calculates the maximum number of components in a graph where each component's sum of values is divisible by k
. The graph is represented as a tree with n
nodes (vertices) and its edges.
Time Complexity:
The time complexity of the function is O(n)
. This is because the function performs a Depth-First Search (DFS) on the tree, starting from node 0
. Each node in the graph is visited only once during the DFS, and at each node, the sum of values is updated and checked for divisibility by k
. Since there are n
nodes and the function visits each node exactly once, the overall time complexity is O(n)
.
Space Complexity:
The space complexity is also O(n)
. This includes:
- The space needed to store the graph
g
, which is an adjacency list representation of the tree. There aren
lists, one for each node, and each list contains the adjacent nodes. Since the tree hasn-1
edges, the total number of elements across all the adjacency lists will be2 * (n - 1)
(each edge is stored twice, once for each node it connects). However, the big-O notation only considers the highest order term, so this is simplified toO(n)
. - The space needed for the DFS recursion call stack. In the worst case, if the tree is implemented as a linked list (i.e., each node has only one child), the maximum depth of the stack would be
n
. Thus, the space complexity due to the call stack isO(n)
as well.
Given that both of these space usages are linear with respect to the number of nodes, the combined space complexity of the function remains O(n)
.
Learn more about how to find time and space complexity quickly using problem constraints.
Which of these pictures shows the visit order of a depth-first search?
Recommended Readings
Everything About Trees A tree is a type of graph data structure composed of nodes and edges Its main properties are It is acyclic doesn't contain any cycles There exists a path from the root to any node Has N 1 edges where N is the number of nodes in the tree and
https algomonster s3 us east 2 amazonaws com cover_photos dfs svg Depth First Search Prereqs Recursion Review problems recursion_intro Trees problems tree_intro With a solid understanding of recursion under our belts we are now ready to tackle one of the most useful techniques in coding interviews Depth First Search DFS
What is Dynamic Programming Prerequisite DFS problems dfs_intro Backtracking problems backtracking Memoization problems memoization_intro Pruning problems backtracking_pruning Dynamic programming is an algorithmic optimization technique that breaks down a complicated problem into smaller overlapping sub problems in a recursive manner and uses solutions to the sub problems to construct a solution
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