LeetCode 2440 - Create Components With Same Value

This problem is asking us to take an undirected tree where each node has a numerical value and determine how many edges we can remove such that every resulting connected component has the same total value.

leetcodehardarraymathtreedepth-first-searchenumeration

LeetCode Problem 2440

Difficulty: 🔴 Hard
Topics: Array, Math, Tree, Depth-First Search, Enumeration

Solution

Problem Understanding

This problem is asking us to take an undirected tree where each node has a numerical value and determine how many edges we can remove such that every resulting connected component has the same total value. The tree is represented by a list of node values nums and a list of edges edges that connect nodes. Deleting an edge splits the tree into two subtrees, and the "value" of a subtree is the sum of the node values in that subtree.

The input guarantees that the graph is a tree, so it has n nodes and n-1 edges, and there is always exactly one path between any two nodes. This is critical because it means that we never have cycles to worry about when summing components, and each split produces independent subtrees.

The output is an integer representing the maximum number of edges that can be deleted while ensuring all components have the same value. Edge cases include a tree with only one node, where no edges exist to delete, and a tree where all node values are equal, which could allow maximum splits.

Constraints indicate that n can go up to 20,000, so an algorithm that explores all possible combinations of edge deletions is too slow. Node values are small (1 <= nums[i] <= 50), which may allow mathematical observations about divisibility.

Important observations include that the sum of the values of all nodes, total_sum, must be divisible by the value of each component we want to create. If total_sum cannot be divided evenly, no valid split is possible.

Approaches

A brute-force approach would attempt to consider every subset of edges to delete, compute the resulting component sums, and check if all sums are equal. While correct in principle, this approach is infeasible because the number of subsets grows exponentially with n.

The key insight for an optimal approach is that the total sum of all nodes must be divisible by the number of components we want to create. If we consider every factor of total_sum as a candidate component value, we can attempt a DFS from the root and track the sum of each subtree. Whenever a subtree matches the candidate component value, it forms a valid component, and we reset its sum for the parent. If all subtrees can be partitioned this way, the number of deletable edges is the number of components minus one.

This approach leverages properties of the tree and simple arithmetic, avoiding exponential search. DFS ensures each node is visited once, making it linear in the number of nodes for each candidate factor.

Approach Time Complexity Space Complexity Notes
Brute Force O(2^n) O(n) Check all subsets of edges for valid splits; infeasible for n up to 20,000
Optimal O(n * sqrt(S)) O(n) DFS for each factor of total sum S, linear traversal per factor

Algorithm Walkthrough

  1. Compute the total sum of all node values, total_sum = sum(nums).
  2. Find all divisors of total_sum because the value of each component must evenly divide total_sum.
  3. For each divisor val, attempt to partition the tree such that every component has value val.
  4. Implement a DFS starting from any node (rooted at 0). For each node, compute the sum of its subtree recursively.
  5. If the sum of a subtree equals val, increment a counter for valid components and return zero to the parent (indicating this subtree is "cut off").
  6. If the sum of a subtree exceeds val, this candidate value is invalid, and we stop the DFS for this factor.
  7. After DFS, if the number of valid components equals total_sum // val, update the maximum number of deletable edges, which is (number of components - 1).
  8. Return the maximum deletable edges among all valid divisors.

Why it works: The algorithm ensures that every subtree sum either matches the candidate component value or propagates up for further aggregation. Because the graph is a tree, each node is counted exactly once, and the DFS guarantees that we can identify every valid component. The use of divisors guarantees that the component sums evenly partition the total sum.

Python Solution

from typing import List
from math import isqrt
from collections import defaultdict

class Solution:
    def componentValue(self, nums: List[int], edges: List[List[int]]) -> int:
        n = len(nums)
        total_sum = sum(nums)
        
        # Build adjacency list
        graph = defaultdict(list)
        for u, v in edges:
            graph[u].append(v)
            graph[v].append(u)
        
        # Function to find all divisors of total_sum
        def divisors(x):
            divs = set()
            for i in range(1, isqrt(x) + 1):
                if x % i == 0:
                    divs.add(i)
                    divs.add(x // i)
            return divs
        
        max_edges = 0
        
        for val in divisors(total_sum):
            target_components = total_sum // val
            found_components = 0
            
            def dfs(node, parent):
                nonlocal found_components
                subtotal = nums[node]
                for nei in graph[node]:
                    if nei == parent:
                        continue
                    child_sum = dfs(nei, node)
                    subtotal += child_sum
                if subtotal == val:
                    found_components += 1
                    return 0
                return subtotal
            
            if dfs(0, -1) == 0 and found_components == target_components:
                max_edges = max(max_edges, found_components - 1)
        
        return max_edges

The Python code first constructs an adjacency list for easy traversal. It computes divisors of the total sum and uses DFS to verify if the tree can be partitioned with that component value. Each subtree that matches the target value counts as a component, and the number of deletable edges is the number of components minus one. This efficiently explores only feasible component values.

Go Solution

package main

func componentValue(nums []int, edges [][]int) int {
    n := len(nums)
    totalSum := 0
    for _, v := range nums {
        totalSum += v
    }

    graph := make([][]int, n)
    for _, e := range edges {
        u, v := e[0], e[1]
        graph[u] = append(graph[u], v)
        graph[v] = append(graph[v], u)
    }

    divisors := func(x int) []int {
        divs := []int{}
        for i := 1; i*i <= x; i++ {
            if x%i == 0 {
                divs = append(divs, i)
                if i != x/i {
                    divs = append(divs, x/i)
                }
            }
        }
        return divs
    }

    maxEdges := 0

    for _, val := range divisors(totalSum) {
        targetComponents := totalSum / val
        foundComponents := 0

        var dfs func(node, parent int) int
        dfs = func(node, parent int) int {
            subtotal := nums[node]
            for _, nei := range graph[node] {
                if nei == parent {
                    continue
                }
                subtotal += dfs(nei, node)
            }
            if subtotal == val {
                foundComponents++
                return 0
            }
            return subtotal
        }

        if dfs(0, -1) == 0 && foundComponents == targetComponents {
            if foundComponents-1 > maxEdges {
                maxEdges = foundComponents - 1
            }
        }
    }

    return maxEdges
}

The Go implementation mirrors Python closely. Slices are used for adjacency lists, and recursion handles subtree sums. Divisors are computed explicitly, and care is taken to avoid duplicate counting of components. Returning zero from DFS corresponds to cutting the subtree.

Worked Examples

Example 1: nums = [6,2,2,2,6], edges = [[0,1],[1,2],[1,3],[3,4]]

Total sum = 18. Divisors: 1, 2, 3, 6, 9, 18.

Candidate val = 6 (18/3 = 3 components):

  • DFS from node 0 returns subtotal 6 -> component found.
  • DFS subtree rooted at node 1 sums to 6 after considering nodes 2 and 3 -> component found.
  • Node 4 alone sums to 6 -> component found.

Number of components = 3, deletable edges = 2.

Example 2: nums = [2], edges = []

Total sum = 2. Only divisor = 1,2. Only one node, cannot delete edges. Result = 0.

Complexity Analysis

Measure Complexity Explanation
Time O(n * sqrt(S)) n nodes visited per DFS, sqrt(S) divisors of total sum S
Space O(n) Adjacency list + recursion stack

DFS is linear in n for each divisor, and divisors of total_sum are bounded by O(sqrt(total_sum)).

Test Cases

# Provided examples
assert Solution().componentValue([6,2,2,2,6], [[0,1],[1,2],[1,3],[3,4]]) == 2  # Example 1
assert Solution().componentValue([2], []) == 0  # Example 2

# Additional cases
assert Solution().componentValue([1,1,1,1], [[