Create Components With Same Value

Create Components With Same Value - Problem

You're given an undirected tree with n nodes labeled from 0 to n-1. Each node has a specific value, and your goal is to strategically delete edges to split the tree into multiple connected components where all components have the same total value.

Input:

nums: An array where nums[i] represents the value of node i
edges: A 2D array where edges[i] = [a, b] represents an edge between nodes a and b

Goal: Return the maximum number of edges you can delete such that every remaining connected component has the same sum of node values.

Key Insight: If we can split the tree into k components with equal values, then each component must have a total value of total_sum / k, where total_sum is the sum of all node values.

Input & Output

example_1.py — Basic Tree Split

$ Input: nums = [6,2,2,2], edges = [[0,1],[1,2],[1,3]]

› Output: 1

💡 Note: We can remove edge [0,1] to create 2 components: {0} with sum 6 and {1,2,3} with sum 6. This is the maximum number of edges we can remove while maintaining equal component sums.

example_2.py — No Valid Split

$ Input: nums = [2], edges = []

› Output: 0

💡 Note: Single node tree cannot be split, so we can remove 0 edges.

example_3.py — Multiple Equal Components

$ Input: nums = [1,1,1,1], edges = [[0,1],[0,2],[0,3]]

› Output: 3

💡 Note: We can remove all 3 edges to create 4 components, each with sum 1. This gives us the maximum number of edge deletions while maintaining equal component sums.

Constraints

1 ≤ nums.length ≤ 2 × 10⁴
1 ≤ nums[i] ≤ 10⁴
edges.length == nums.length - 1
edges[i].length == 2
0 ≤ edges[i][0], edges[i][1] < nums.length
The input represents a valid tree

Visualization

Tap to expand

Asked in

G Google 45 a Amazon 38 f Meta 32 ⊞ Microsoft 28

The optimal solution uses DFS with target sum enumeration. We try each possible number of components k, calculate the required sum per component (total_sum/k), then use DFS to count how many edges can be cut while maintaining this constraint. The key insight is that an edge can only be cut if the subtree below it has exactly the target sum, allowing it to become an independent component.

Common Approaches

✓ Stack

⏱️ Time: N/A Space: N/A

Brute Force (Try All Edge Combinations)

⏱️ Time: O(2^n * n^2) Space: O(n^2)

Generate all possible combinations of edges to delete, check if resulting components have equal sums. This approach has exponential time complexity as there are 2^(n-1) possible combinations.

DFS with Target Sum Enumeration (Optimal)

⏱️ Time: O(n^2) Space: O(n)

Instead of trying all edge combinations, we enumerate possible target sums for each component. For a valid split into k components, each must have sum = total_sum/k. We use DFS to count how many edges we can cut while maintaining this constraint.

Algorithm Steps — Algorithm Steps

Code -

solution.c — C

Time & Space Complexity

Time Complexity

⏱️

⚠ Quadratic Growth

Space Complexity

⚠ Quadratic Space

43.7K Views

Medium Frequency

~35 min Avg. Time

1.8K Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen