What is the time complexity of Create Components With Same Value?

The optimal solution has a time complexity of O(n*m).

What data structures are used to solve Create Components With Same Value?

This problem uses Array, Math, Tree, Depth-First Search, Enumeration. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Create Components With Same Value?

Insight 1: The number of components, `k`, must be a divisor of the total sum of all node values. This is because the total sum must be evenly divisible by the component sum to form `k` components each having that same sum. Insight 2: We can use Depth-First Search (DFS) to traverse the tree and calculate the sum of node values for each subtree. We can determine if a subtree's sum equals the target component value by strategically 'cutting' the edge between the subtree and its parent.

Which companies ask LeetCode 2440. Create Components With Same Value in interviews?

This problem is frequently asked at Sprinklr.

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Create Components With Same Value - LeetCode 2440 Solution

Create Components With Same Value - Complete Solution Guide

Create Components With Same Value is LeetCode problem 2440, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

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 i th node. You are also given a 2D integer array edges of length n - 1 where edges[i] = [a i , b i ] indicates that there is an edge between nodes a i and b i 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

Detailed Explanation

The problem asks us to find the maximum number of edges that can be removed from a given undirected tree such that the remaining connected components all have the same sum of node values. We are given an array `nums` representing the values of each node and a 2D array `edges` representing the connections between nodes. The goal is to split the tree into components, each having the same value, by removing edges.

Solution Approach

The solution iterates through the possible number of components (`k`), starting from `n` down to 2. For each `k`, it checks if the total sum of the `nums` array is divisible by `k`. If it is, it calculates the target sum for each component (total_sum / k) and then uses Depth-First Search (DFS) to try to partition the tree into components with the target sum. The DFS function recursively calculates the sum of each subtree. If a subtree's sum equals the target sum, it effectively 'cuts' the edge connecting that subtree to its parent, forming a valid component. The DFS returns 0 if the entire tree can be successfully partitioned into `k` components each with a sum equal to `target`. If this is the case, it means we can delete k-1 edges. If no such partition can be found for any k > 1, the answer will be 0 which is the case where the entire tree is a single component and no edges are deleted.

Step-by-Step Algorithm

Step 1: Calculate the total sum of the `nums` array.
Step 2: Iterate `k` from `n` down to 2. `k` represents the potential number of components.
Step 3: Inside the loop, check if `total_sum` is divisible by `k`. If not, continue to the next iteration.
Step 4: If `total_sum` is divisible by `k`, calculate `target = total_sum / k`. `target` is the desired sum of each component.
Step 5: Define a DFS function that takes the current node `u`, the parent node `p`, and the `target` sum as input.
Step 6: Inside the DFS function, calculate the `current_sum` of the subtree rooted at `u`.
Step 7: Iterate through the neighbors `v` of `u`. If `v` is the parent `p`, skip it.
Step 8: Recursively call DFS on neighbor `v` to get the `child_sum`.
Step 9: If `child_sum` is -1, it means a valid partition could not be found in the subtree rooted at v, so propagate the -1 up.
Step 10: Add the `child_sum` to the `current_sum`.
Step 11: After processing all children, check the `current_sum`. If it's greater than `target`, return -1 (invalid partition). If it's equal to `target`, return 0 (subtree can be considered a component). Otherwise, return `current_sum` to be added to the parent's sum.
Step 12: Call the DFS function starting from node 0 (root) with parent -1 and the calculated `target`.
Step 13: If the DFS returns 0, it means the entire tree can be partitioned into components with sum `target`, so return `k - 1`.
Step 14: If the loop completes without finding a valid `k`, return 0 (no edges can be deleted).

Key Insights

Insight 1: The number of components, `k`, must be a divisor of the total sum of all node values. This is because the total sum must be evenly divisible by the component sum to form `k` components each having that same sum.
Insight 2: We can use Depth-First Search (DFS) to traverse the tree and calculate the sum of node values for each subtree. We can determine if a subtree's sum equals the target component value by strategically 'cutting' the edge between the subtree and its parent.
Insight 3: To maximize the number of deleted edges, we want to maximize the number of components, `k`. Therefore, we should iterate through potential values of `k` (number of components) from `n` (the number of nodes) down to 1, and check if a valid partition exists for each `k`.

Complexity Analysis

Time Complexity: O(n*m)

Space Complexity: O(n)

Topics

This problem involves: Array, Math, Tree, Depth-First Search, Enumeration.

Companies

Asked at: Sprinklr.

Create Components With Same Value | LeetCode 2440

Hard

Array Math Tree Depth-First Search Enumeration

Time: O(n*m)Space: O(n)

Solution Code

Available in:

import java.util.ArrayList;
import java.util.List;

class Solution {
    public int componentValue(int[] nums, int[][] edges) {
        int n = nums.length;
        if (n == 1) {
            return 0;
        }

        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }

        for (int[] edge : edges) {
            int u = edge[0];
            int v = edge[1];
            adj.get(u).add(v);
            adj.get(v).add(u);
        }

        int totalSum = 0;
        for (int num : nums) {
            totalSum += num;
        }

        class DFSResult {
            int result;

            public DFSResult(int result) {
                this.result = result;
            }
        }

        class DFSHelper {
            int dfs(int u, int p, int target, List<List<Integer>> adj, int[] nums) {
                int currentSum = nums[u];
                for (int v : adj.get(u)) {
                    if (v == p) {
                        continue;
                    }

                    int childSum = dfs(v, u, target, adj, nums);

                    if (childSum == -1) {
                        return -1;
                    }

                    currentSum += childSum;
                }

                if (currentSum > target) {
                    return -1;
                } else if (currentSum == target) {
                    return 0;
                } else {
                    return currentSum;
                }
            }
        }

        DFSHelper dfsHelper = new DFSHelper();


        for (int k = n; k > 1; k--) {
            if (totalSum % k == 0) {
                int target = totalSum / k;
                if (dfsHelper.dfs(0, -1, target, adj, nums) == 0) {
                    return k - 1;
                }
            }
        }

        return 0;
    }
}

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Problem Info

Description

Problem 2440: Create Components With Same Value - Detailed explanation and solution approaches.

Topics

Array Math Tree Depth-First Search Enumeration

Companies

Sprinklr

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2440

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

14 steps

1
Step 1: Calculate the total sum of the `nums` array.
2
Step 2: Iterate `k` from `n` down to 2. `k` represents the potential number of components.
3
Step 3: Inside the loop, check if `total_sum` is divisible by `k`. If not, continue to the next iteration.
4
Step 4: If `total_sum` is divisible by `k`, calculate `target = total_sum / k`. `target` is the desired sum of each component.
5
Step 5: Define a DFS function that takes the current node `u`, the parent node `p`, and the `target` sum as input.
6
Step 6: Inside the DFS function, calculate the `current_sum` of the subtree rooted at `u`.
7
Step 7: Iterate through the neighbors `v` of `u`. If `v` is the parent `p`, skip it.
8
Step 8: Recursively call DFS on neighbor `v` to get the `child_sum`.
9
Step 9: If `child_sum` is -1, it means a valid partition could not be found in the subtree rooted at v, so propagate the -1 up.
10
Step 10: Add the `child_sum` to the `current_sum`.
11
Step 11: After processing all children, check the `current_sum`. If it's greater than `target`, return -1 (invalid partition). If it's equal to `target`, return 0 (subtree can be considered a component). Otherwise, return `current_sum` to be added to the parent's sum.
12
Step 12: Call the DFS function starting from node 0 (root) with parent -1 and the calculated `target`.
13
Step 13: If the DFS returns 0, it means the entire tree can be partitioned into components with sum `target`, so return `k - 1`.
14
Step 14: If the loop completes without finding a valid `k`, return 0 (no edges can be deleted).

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n*m)

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is derived from iterating through possible numbers of components and performing a DFS for each. The space complexity comes from storing the graph in the adjacency list and the recursion stack during the DFS traversal.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Dynamic Programming

Instead of DFS, we could potentially use dynamic programming to solve this problem. However, because the tree is undirected, using standard tree DP can be complex. You'd need to consider states representing subtrees and their possible sums, which might increase the space complexity.

Time: Potentially O(n*sum), where sum is the total sum of all node values.Space: Potentially O(n*sum)

Trade-offs: DP might be useful if memory is not a constraint and more complex pruning can be applied. DFS is typically more straightforward for tree problems.

✨

Key Insights

1Insight 1: The number of components, `k`, must be a divisor of the total sum of all node values. This is because the total sum must be evenly divisible by the component sum to form `k` components each having that same sum.
2Insight 2: We can use Depth-First Search (DFS) to traverse the tree and calculate the sum of node values for each subtree. We can determine if a subtree's sum equals the target component value by strategically 'cutting' the edge between the subtree and its parent.
3Insight 3: To maximize the number of deleted edges, we want to maximize the number of components, `k`. Therefore, we should iterate through potential values of `k` (number of components) from `n` (the number of nodes) down to 1, and check if a valid partition exists for each `k`.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: `n = 1`. If there is only one node, there are no edges to delete, so the answer is 0. The code handles this case explicitly.
Edge case 2: The total sum is a prime number. In this case, the only possible values of `k` are 1 and the total sum itself. Since we want `k > 1`, only the case where the whole tree is one component will be considered, returning 0.
Edge case 3: Deep trees that can exceed python's default recursion limit which the solution tries to address by increasing it.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not considering all possible divisors of the total sum. The number of components must be a divisor of the total sum, and the solution needs to iterate through these divisors.

✗Common mistake 2: Incorrectly calculating the sum of subtrees in the DFS. The DFS function needs to accurately track the sum of the current subtree and cut the edge if the sum matches the target component value.

✗Common mistake 3: Not handling the base case of n=1 correctly.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network partitioning. This algorithm can be adapted to partition a network into subnetworks with equal resource allocation (e.g., bandwidth or processing power), maximizing the number of isolated subnetworks.

🏦Real-world scenario 2: Task scheduling. Consider a set of tasks with dependencies represented by a tree. Each task has a cost (nums[i]). The problem can be adapted to determine the maximum number of independent task groups with equal total cost that can be executed concurrently.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: September 9, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Create Components With Same Value"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?