LeetCode #2440 — HARD

Create Components With Same Value

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

Solve on LeetCode
The Problem

Problem Statement

There is an undirected tree with n nodes labeled from 0 to n - 1.

You are given a 0-indexed integer array nums of length n where nums[i] represents the value of the ith node. You are also given a 2D integer array edges of length n - 1 where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree.

You are allowed to delete some edges, splitting the tree into multiple connected components. Let the value of a component be the sum of all nums[i] for which node i is in the component.

Return the maximum number of edges you can delete, such that every connected component in the tree has the same value.

Example 1:

Input: nums = [6,2,2,2,6], edges = [[0,1],[1,2],[1,3],[3,4]] 
Output: 2 
Explanation: The above figure shows how we can delete the edges [0,1] and [3,4]. The created components are nodes [0], [1,2,3] and [4]. The sum of the values in each component equals 6. It can be proven that no better deletion exists, so the answer is 2.

Example 2:

Input: nums = [2], edges = []
Output: 0
Explanation: There are no edges to be deleted.

Constraints:

  • 1 <= n <= 2 * 104
  • nums.length == n
  • 1 <= nums[i] <= 50
  • edges.length == n - 1
  • edges[i].length == 2
  • 0 <= edges[i][0], edges[i][1] <= n - 1
  • edges represents a valid tree.
Patterns Used
🌳Tree Traversal→

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: There is an undirected tree with n nodes labeled from 0 to n - 1. You are given a 0-indexed integer array nums of length n where nums[i] represents the value of the ith node. You are also given a 2D integer array edges of length n - 1 where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the tree. You are allowed to delete some edges, splitting the tree into multiple connected components. Let the value of a component be the sum of all nums[i] for which node i is in the component. Return the maximum number of edges you can delete, such that every connected component in the tree has the same value.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Math · Tree

Example 1

[6,2,2,2,6]
[[0,1],[1,2],[1,3],[3,4]]

Example 2

[2]
[]

Related Problems

  • Equal Tree Partition (equal-tree-partition)
  • Maximum Number of K-Divisible Components (maximum-number-of-k-divisible-components)
Step 02

Core Insight

What unlocks the optimal approach

  • Consider all divisors of the sum of values.
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Largest constraint values
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2440: Create Components With Same Value
class Solution {
    private List<Integer>[] g;
    private int[] nums;
    private int t;

    public int componentValue(int[] nums, int[][] edges) {
        int n = nums.length;
        g = new List[n];
        this.nums = nums;
        Arrays.setAll(g, k -> new ArrayList<>());
        for (var e : edges) {
            int a = e[0], b = e[1];
            g[a].add(b);
            g[b].add(a);
        }
        int s = sum(nums), mx = max(nums);
        for (int k = Math.min(n, s / mx); k > 1; --k) {
            if (s % k == 0) {
                t = s / k;
                if (dfs(0, -1) == 0) {
                    return k - 1;
                }
            }
        }
        return 0;
    }

    private int dfs(int i, int fa) {
        int x = nums[i];
        for (int j : g[i]) {
            if (j != fa) {
                int y = dfs(j, i);
                if (y == -1) {
                    return -1;
                }
                x += y;
            }
        }
        if (x > t) {
            return -1;
        }
        return x < t ? x : 0;
    }

    private int sum(int[] arr) {
        int x = 0;
        for (int v : arr) {
            x += v;
        }
        return x;
    }

    private int max(int[] arr) {
        int x = arr[0];
        for (int v : arr) {
            x = Math.max(x, v);
        }
        return x;
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n)
Space
O(h)

Approach Breakdown

LEVEL ORDER
O(n) time
O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL
O(n) time
O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.

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