How do you solve Create Components With Same Value on LeetCode?

Create Components With Same Value (LeetCode #2440) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding the relationship between the total sum of node values and the components formed is crucial.

What is the brute force solution for Create Components With Same Value?

The brute force approach for Create Components With Same Value is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking all possible ways to delete edges and see if the resulting components have the same sum. This is straightforward but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Create Components With Same Value?

Create Components With Same Value uses the following concepts: Array, Math, Tree, Depth-First Search, Enumeration. The recommended patterns to study are: DFS for tree traversal., Graph representation using adjacency lists..

What are the interview tips for Create Components With Same Value?

Always clarify the problem constraints and edge cases. Think about how to break down the problem into smaller parts. Practice explaining your thought process as you code.

What are common mistakes when solving Create Components With Same Value?

Failing to consider all divisors of the total sum. Not correctly implementing the DFS to track visited nodes.

#2440

Create Components With Same Value

Hard

ArrayMathTreeDepth-First SearchEnumerationDFS for tree traversal.Graph representation using adjacency lists.

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach leverages the fact that we only need to consider the divisors of the total sum of the node values. By using a DFS to calculate component sums, we can efficiently determine how many edges can be deleted.

⚙️

Algorithm

4 steps

1Step 1: Calculate the total sum of the nums array.
2Step 2: Find all divisors of the total sum.
3Step 3: For each divisor, use DFS to check how many components can be formed with that sum.
4Step 4: Keep track of the maximum number of edges that can be deleted for valid components.

solution.py35 lines

1# Full working Python code
2from collections import defaultdict
3
4def maxEdgesToDelete(nums, edges):
5    total_sum = sum(nums)
6    n = len(nums)
7    graph = defaultdict(list)
8    for a, b in edges:
9        graph[a].append(b)
10        graph[b].append(a)
11    
12    def dfs(node, parent):
13        component_sum = nums[node]
14        for neighbor in graph[node]:
15            if neighbor != parent:
16                component_sum += dfs(neighbor, node)
17        return component_sum
18    
19    max_deletions = 0
20    for divisor in range(1, total_sum + 1):
21        if total_sum % divisor == 0:
22            current_sum = 0
23            deletions = 0
24            visited = [False] * n
25            for i in range(n):
26                if not visited[i]:
27                    component_sum = dfs(i, -1)
28                    if component_sum == divisor:
29                        deletions += 1
30                    current_sum += component_sum
31                    visited[i] = True
32            if current_sum == total_sum:
33                max_deletions = max(max_deletions, deletions - 1)
34    return max_deletions
35

ℹ

Complexity note: The time complexity is O(n) as we only traverse the tree once for each divisor, and the space complexity is O(n) due to the graph representation.

1Understanding the relationship between the total sum of node values and the components formed is crucial.
2Divisors of the total sum guide the possible valid component sums.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.