What is the time complexity of Maximize Sum of Weights after Edge Removals?

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

What data structures are used to solve Maximize Sum of Weights after Edge Removals?

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

What are the key insights for solving Maximize Sum of Weights after Edge Removals?

Insight 1: The problem can be solved using dynamic programming on trees. We can define a state for each node that considers whether the edge connecting the node to its parent is kept or removed. Insight 2: The optimal decision for a node depends on the optimal decisions made for its children. This allows us to use a bottom-up approach via Depth-First Search (DFS).

Which companies ask LeetCode 3367. Maximize Sum of Weights after Edge Removals in interviews?

This problem is frequently asked at Gameskraft.

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Maximize Sum of Weights after Edge Removals - LeetCode 3367 Solution

Maximize Sum of Weights after Edge Removals - Complete Solution Guide

Maximize Sum of Weights after Edge Removals is LeetCode problem 3367, 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 exists an undirected tree with n nodes numbered 0 to n - 1 . You are given a 2D integer array edges of length n - 1 , where edges[i] = [u i , v i , w i ] indicates that there is an edge between nodes u i and v i with weight w i in the tree. Your task is to remove zero or more edges such that: Each node has an edge with at most k other nodes, where k is given. The sum of the weights of the remaining edges is maximized . Return the maximum possible sum of weights for the remaining edges afte

Detailed Explanation

The problem asks us to find the maximum sum of edge weights in a given undirected tree after removing some edges. The constraint is that after the removals, each node in the tree must be connected to at most 'k' other nodes. Essentially, we need to strategically remove edges to satisfy the degree constraint while maximizing the total weight of the remaining edges.

Solution Approach

The solution uses dynamic programming with Depth-First Search (DFS) to traverse the tree. For each node, we consider two states: whether the edge connecting the node to its parent is removed (state 0) or kept (state 1). The DFS function calculates the maximum weight that can be obtained from the subtree rooted at a node, considering these two states. It explores each child, determines the benefit of keeping the edge to the child versus removing it, sorts these benefits, and then picks the top 'k' (or 'k-1' if we consider keeping the parent edge as well) to maximize the total weight. Memoization is employed to avoid redundant computations.

Step-by-Step Algorithm

Step 1: Build the adjacency list representation of the tree from the given 'edges' array.
Step 2: Define a recursive DFS function that takes a node 'u' and its parent 'p' as input.
Step 3: Inside the DFS function, initialize a list 'gains' to store the differences between keeping and removing edges to the children of 'u'. Also, initialize 'base_sum' to accumulate the weights if all edges to children are removed.
Step 4: Iterate through the neighbors 'v' of 'u'. If 'v' is not the parent 'p', recursively call the DFS function for 'v'.
Step 5: Calculate the weight gained by keeping the edge to 'v' versus removing it. This involves calculating `v_keep_val` (weight of the edge + result from dfs(v,u) when edge to parent v is included) and `v_remove_val` (result from dfs(v,u) when edge to parent v is not included)
Step 6: Add the difference (`v_keep_val - v_remove_val`) to the 'gains' list if it's positive, otherwise just add `v_remove_val` to `base_sum`
Step 7: Sort the 'gains' list in descending order.
Step 8: Calculate `dp_u_0` (maximum weight if the edge to the parent is removed) by adding the top 'k' gains to 'base_sum'.
Step 9: Calculate `dp_u_1` (maximum weight if the edge to the parent is kept) by adding the top 'k-1' gains to 'base_sum', only if k > 0.
Step 10: Store the results (`dp_u_0`, `dp_u_1`) in the memoization table.
Step 11: Return the result of DFS starting from an arbitrary root node (e.g., node 0).

Key Insights

Insight 1: The problem can be solved using dynamic programming on trees. We can define a state for each node that considers whether the edge connecting the node to its parent is kept or removed.
Insight 2: The optimal decision for a node depends on the optimal decisions made for its children. This allows us to use a bottom-up approach via Depth-First Search (DFS).
Insight 3: For each node, we need to choose at most 'k' children to keep the edges connected to them. This involves sorting the 'gains' (the difference between keeping and removing an edge) and selecting the top 'k' gains.

Complexity Analysis

Time Complexity: O(n*log(n))

Space Complexity: O(n)

Topics

This problem involves: Dynamic Programming, Tree, Depth-First Search.

Companies

Asked at: Gameskraft.

Maximize Sum of Weights after Edge Removals | LeetCode 3367

Hard

Dynamic Programming Tree Depth-First Search

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

Solution Code

Available in:

import java.util.*;

class Solution {
    public long maximizeSumOfWeights(int[][] edges, int k) {
        int n = edges.length + 1;
        if (n <= 1) {
            return 0;
        }

        List<List<Pair>> 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];
            int w = edge[2];
            adj.get(u).add(new Pair(v, w));
            adj.get(v).add(new Pair(u, w));
        }

        Map<String, Long[]> memo = new HashMap<>();

        function dfs = new function() {
            public Long[] apply(int u, int p) {
                String key = u + "," + p;
                if (memo.containsKey(key)) {
                    return memo.get(key);
                }

                List<Long> gains = new ArrayList<>();
                long baseSum = 0;

                for (Pair pair : adj.get(u)) {
                    int v = pair.node;
                    int w = pair.weight;
                    if (v == p) {
                        continue;
                    }

                    Long[] dp_v = dfs.apply(v, u);
                    long dp_v_0 = dp_v[0];
                    long dp_v_1 = dp_v[1];

                    long v_remove_val = dp_v_0;
                    long v_keep_val = w + dp_v_1;

                    baseSum += v_remove_val;
                    if (v_keep_val > v_remove_val) {
                        gains.add(v_keep_val - v_remove_val);
                    }
                }

                Collections.sort(gains, Collections.reverseOrder());

                long dp_u_0 = baseSum;
                for (int i = 0; i < Math.min(k, gains.size()); i++) {
                    dp_u_0 += gains.get(i);
                }

                long dp_u_1 = baseSum;
                if (k > 0) {
                    for (int i = 0; i < Math.min(k - 1, gains.size()); i++) {
                        dp_u_1 += gains.get(i);
                    }
                }

                Long[] result = new Long[]{dp_u_0, dp_u_1};
                memo.put(key, result);
                return result;
            }
        };

        Long[] result = dfs.apply(0, -1);
        return result[0];
    }

    static class Pair {
        int node;
        int weight;

        public Pair(int node, int weight) {
            this.node = node;
            this.weight = weight;
        }
    }

    interface function {
        Long[] apply(int u, int p);
    }
}

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Problem Info

Description

Problem 3367: Maximize Sum of Weights after Edge Removals - Detailed explanation and solution approaches.

Topics

Dynamic Programming Tree Depth-First Search

Companies

Gameskraft

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3367

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

11 steps

1
Step 1: Build the adjacency list representation of the tree from the given 'edges' array.
2
Step 2: Define a recursive DFS function that takes a node 'u' and its parent 'p' as input.
3
Step 3: Inside the DFS function, initialize a list 'gains' to store the differences between keeping and removing edges to the children of 'u'. Also, initialize 'base_sum' to accumulate the weights if all edges to children are removed.
4
Step 4: Iterate through the neighbors 'v' of 'u'. If 'v' is not the parent 'p', recursively call the DFS function for 'v'.
5
Step 5: Calculate the weight gained by keeping the edge to 'v' versus removing it. This involves calculating `v_keep_val` (weight of the edge + result from dfs(v,u) when edge to parent v is included) and `v_remove_val` (result from dfs(v,u) when edge to parent v is not included)
6
Step 6: Add the difference (`v_keep_val - v_remove_val`) to the 'gains' list if it's positive, otherwise just add `v_remove_val` to `base_sum`
7
Step 7: Sort the 'gains' list in descending order.
8
Step 8: Calculate `dp_u_0` (maximum weight if the edge to the parent is removed) by adding the top 'k' gains to 'base_sum'.
9
Step 9: Calculate `dp_u_1` (maximum weight if the edge to the parent is kept) by adding the top 'k-1' gains to 'base_sum', only if k > 0.
10
Step 10: Store the results (`dp_u_0`, `dp_u_1`) in the memoization table.
11
Step 11: Return the result of DFS starting from an arbitrary root node (e.g., node 0).

💡

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*log(n))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The time complexity is determined by the DFS traversal and the sorting of gains. The DFS itself is O(n). The sorting of gains for each node contributes O(n log n) overall. The space complexity comes from the adjacency list, the memoization table, and the recursion stack.

📖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

Iterative Dynamic Programming

Instead of using recursion, we could implement the dynamic programming solution iteratively. This would involve a topological sort of the tree and then iterating through the nodes in reverse topological order to calculate the DP values.

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

Trade-offs: Iterative DP avoids the recursion stack overhead, which could be beneficial for very large trees. However, it might be more complex to implement than the recursive approach.

✨

Key Insights

1Insight 1: The problem can be solved using dynamic programming on trees. We can define a state for each node that considers whether the edge connecting the node to its parent is kept or removed.
2Insight 2: The optimal decision for a node depends on the optimal decisions made for its children. This allows us to use a bottom-up approach via Depth-First Search (DFS).
3Insight 3: For each node, we need to choose at most 'k' children to keep the edges connected to them. This involves sorting the 'gains' (the difference between keeping and removing an edge) and selecting the top 'k' gains.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Edge Cases

Edge case 1: Empty tree (n <= 1). The code handles this by returning 0 immediately.
Edge case 2: k = 0. In this case, we must remove all edges. The algorithm correctly handles this by effectively not adding any 'gains' when calculating dp_u_0 and dp_u_1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Forgetting to memoize the results of the DFS function, which would lead to exponential time complexity.

✗Common mistake 2: Incorrectly calculating the 'gains' or not sorting them properly before selecting the top 'k' values.

✗Common mistake 3: Not handling the case where k = 0 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 design. Consider a communication network where nodes represent devices and edges represent communication links. We might want to maximize the bandwidth of the network (sum of edge capacities) while ensuring that each device is connected to at most 'k' other devices (due to hardware limitations or security policies).

🏦Real-world scenario 2: Social network analysis. Imagine a social network where nodes are users and edges represent connections. We might want to find a sub-network that maximizes the total 'influence' (represented by edge weights) while limiting the number of connections each user can have (to prevent spam or maintain privacy).

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

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

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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: October 6, 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

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJul 13

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 6

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJun 29

❓

Frequently Asked Questions

Should I use top-down (memoization) or bottom-up (tabulation)?

Start with top-down (recursion + memoization) as it directly maps to the recursive thinking and handles only needed subproblems. Convert to bottom-up for better space efficiency and to avoid stack overflow. Bottom-up is often faster due to no recursion overhead.

What is the difference between DFS and BFS for trees?

DFS (using recursion or stack) explores depth-first, useful for path-finding and backtracking problems. BFS (using queue) explores level-by-level, ideal for finding shortest paths or processing nodes by depth. Choose based on whether level information matters.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximize Sum of Weights after Edge Removals"?

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

🤖 Need help with this problem?