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Create Components With Same Value

Number: 2531

Difficulty: Hard

Paid? No

Companies: Sprinklr

Problem Description

Given an undirected tree with n nodes (labeled from 0 to n-1) where each node has an assigned value from the array nums, and a list of edges describing the tree, you are allowed to delete some of the edges. This deletion splits the tree into connected components. The value of a component is defined as the sum of the nums values for all nodes within that component. The goal is to delete as many edges as possible such that every resulting connected component has the same total value.

Key Insights

  • The overall sum of nums must be divisible by the candidate component value.
  • A valid candidate component sum must be a divisor of the total sum.
  • Using DFS, you can accumulate subtree sums; when a subtree sum exactly equals the candidate, you “cut” the subtree and reset its sum to 0.
  • The maximum number of deletions is the number of valid components minus one.
  • Attempt all divisors (except the trivial case where there is one component) and select the one that yields the maximum deletions.

Space and Time Complexity

Time Complexity: O(n * d) where n is the number of nodes and d is the number of divisors of the total sum. Space Complexity: O(n) due to recursion stack (in worst-case skewed tree) and the adjacency list.

Solution

We start by calculating the total sum of all node values. For each divisor of the total sum (candidate component sum), we try to partition the tree into components where each component sums exactly to the candidate. We perform a DFS starting from an arbitrary root. At each node, we sum up the values of its subtree. Once the sum becomes exactly equal to the candidate, it means we have a valid component and we “cut” this subtree (return 0 to the parent) while counting this as one valid component. If the accumulated sum exceeds the candidate, the candidate is invalid for partitioning. After processing all candidates, the answer will be the maximum number of deletions (number of components formed minus one).

Key data structures and techniques include:

  • An adjacency list to represent the tree.
  • DFS recursion for accumulating and checking subtree sums.
  • Iteration over the divisors of the total sum.

Code Solutions

# Python solution with detailed comments

class Solution:
    def componentValue(self, nums, edges):
        # Build tree as an adjacency list
        n = len(nums)
        tree = [[] for _ in range(n)]
        for u, v in edges:
            tree[u].append(v)
            tree[v].append(u)
        
        total_sum = sum(nums)
        
        # Function to perform DFS, returns the subtree sum modulo candidate
        def dfs(node, parent, candidate):
            current_sum = nums[node]
            # Iterate through children (neighbors except the parent)
            for child in tree[node]:
                if child != parent:
                    sub_sum = dfs(child, node, candidate)
                    if sub_sum == -1:  # propagate failure
                        return -1
                    current_sum += sub_sum
            # If the subtree sum equals candidate, treat it as forming a valid component
            if current_sum == candidate:
                # Return 0 to indicate that we've cut here and do not add to parent's sum
                self.count += 1  # increment count for a complete valid component
                return 0
            # If current sum exceeds candidate, the candidate is invalid so return failure signal
            if current_sum > candidate:
                return -1
            # Otherwise, return the accumulated sum so far
            return current_sum
        
        max_deletions = 0
        
        # Enumerate each possible candidate value (divisor of total_sum)
        # The candidate should be less than total_sum to have at least 2 components.
        for candidate in range(1, total_sum):
            if total_sum % candidate != 0:
                continue  # candidate must divide total_sum evenly
            self.count = 0  # initialize count of valid components
            if dfs(0, -1, candidate) == 0:
                # Check if we formed exactly the expected number of components
                components = self.count
                if components > 1:  # At least two components means an edge was deleted
                    max_deletions = max(max_deletions, components - 1)
        return max_deletions

# Example usage:
# sol = Solution()
# print(sol.componentValue([6,2,2,2,6], [[0,1],[1,2],[1,3],[3,4]]))
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