What is the time complexity of Find the Maximum Sum of Node Values?

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

What data structures are used to solve Find the Maximum Sum of Node Values?

This problem uses Array, Dynamic Programming, Greedy, Bit Manipulation, Tree, Sorting. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Find the Maximum Sum of Node Values?

Insight 1: The key insight is that performing the XOR operation twice on the same edge restores the original values. This means we only need to consider performing the operation zero or one time on each edge. Insight 2: The parity (even or odd) of the number of nodes XORed with `k` matters. If an even number of nodes have been XORed, the total sum is not affected (as each XOR is paired up). If an odd number of nodes have been XORed with `k`, we should minimize the loss from the XOR operations by XORing the value that results in the least overall difference (absolute difference between the original value and the XORed value).

Which companies ask LeetCode 3068. Find the Maximum Sum of Node Values in interviews?

This problem is frequently asked at BlackRock, Deutsche Bank.

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Find the Maximum Sum of Node Values - LeetCode 3068 Solution

Find the Maximum Sum of Node Values - Complete Solution Guide

Find the Maximum Sum of Node Values is LeetCode problem 3068, 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 0-indexed 2D integer array edges of length n - 1 , where edges[i] = [u i , v i ] indicates that there is an edge between nodes u i and v i in the tree. You are also given a positive integer k , and a 0-indexed array of non-negative integers nums of length n , where nums[i] represents the value of the node numbered i . Alice wants the sum of values of tree nodes to be maximum , for which Alice can perform the follo

Detailed Explanation

The problem describes a tree structure where each node has a value. We are given an array `nums` representing the values of the nodes, a value `k`, and an array `edges` representing the connections between nodes in the tree. The goal is to maximize the sum of the node values by performing XOR operations on connected nodes. Specifically, we can choose any edge (u, v) and perform `nums[u] = nums[u] XOR k` and `nums[v] = nums[v] XOR k`. We can perform this operation any number of times, including zero. The output is the maximum possible sum of the node values.

Solution Approach

The solution iterates through each node and determines whether XORing it with `k` would increase its value. It keeps track of the sum of the maximum values achievable for each node (either the original value or the XORed value). Also, it counts the number of nodes that are flipped and finds the minimum absolute difference between the original and flipped value for all nodes. Finally, if the number of flipped nodes is odd, the solution subtracts the minimum loss from the sum, as an odd number of flips means there's no way to 'cancel out' all XORs with another XOR, and the optimal action requires reducing overall sum by smallest value that was reduced by the XOR operation.

Step-by-Step Algorithm

Step 1: Initialize `max_sum` to 0, `flipped_count` to 0, and `min_loss` to infinity.
Step 2: Iterate through each node value `num` in `nums`.
Step 3: Calculate `flipped_num = num XOR k`.
Step 4: If `flipped_num > num`, add `flipped_num` to `max_sum` and increment `flipped_count`. Otherwise, add `num` to `max_sum`.
Step 5: Calculate the absolute difference `loss = abs(num - flipped_num)` and update `min_loss = min(min_loss, loss)`.
Step 6: After iterating through all nodes, check if `flipped_count` is even. If it is, return `max_sum`. Otherwise, return `max_sum - min_loss`.

Key Insights

Insight 1: The key insight is that performing the XOR operation twice on the same edge restores the original values. This means we only need to consider performing the operation zero or one time on each edge.
Insight 2: The parity (even or odd) of the number of nodes XORed with `k` matters. If an even number of nodes have been XORed, the total sum is not affected (as each XOR is paired up). If an odd number of nodes have been XORed with `k`, we should minimize the loss from the XOR operations by XORing the value that results in the least overall difference (absolute difference between the original value and the XORed value).
Insight 3: The tree structure itself doesn't fundamentally impact the approach other than the XOR operations apply between adjacent nodes. No tree traversal is required

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Dynamic Programming, Greedy, Bit Manipulation, Tree, Sorting.

Companies

Asked at: BlackRock, Deutsche Bank.

Find the Maximum Sum of Node Values | LeetCode 3068

Hard

Array Dynamic Programming Greedy Bit Manipulation Tree Sorting

Time: O(n)Space: O(1)

Solution Code

Available in:

class Solution {
    public long maximumValueSum(int[] nums, int k, int[][] edges) {
        long max_sum = 0;
        int flipped_count = 0;
        long min_loss = Integer.MAX_VALUE;

        for (int num : nums) {
            int flipped_num = num ^ k;
            if (flipped_num > num) {
                max_sum += flipped_num;
                flipped_count++;
            } else {
                max_sum += num;
            }

            long loss = Math.abs(num - flipped_num);
            min_loss = Math.min(min_loss, loss);
        }

        if (flipped_count % 2 == 0) {
            return max_sum;
        } else {
            return max_sum - min_loss;
        }
    }
}

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Problem Info

Description

Problem 3068: Find the Maximum Sum of Node Values - Detailed explanation and solution approaches.

Topics

Array Dynamic Programming Greedy Bit Manipulation Tree Sorting

Companies

BlackRockDeutsche Bank

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:3068

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

6 steps

1
Step 1: Initialize `max_sum` to 0, `flipped_count` to 0, and `min_loss` to infinity.
2
Step 2: Iterate through each node value `num` in `nums`.
3
Step 3: Calculate `flipped_num = num XOR k`.
4
Step 4: If `flipped_num > num`, add `flipped_num` to `max_sum` and increment `flipped_count`. Otherwise, add `num` to `max_sum`.
5
Step 5: Calculate the absolute difference `loss = abs(num - flipped_num)` and update `min_loss = min(min_loss, loss)`.
6
Step 6: After iterating through all nodes, check if `flipped_count` is even. If it is, return `max_sum`. Otherwise, return `max_sum - min_loss`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the single loop that iterates through all n nodes. The space complexity is constant because the amount of extra memory used does not depend on the input size.

📖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

While the problem mentions Dynamic Programming in the topics, the given solution and constraints of the problem allow for a simpler greedy approach. A DP approach would likely involve defining a state based on the node index and whether it's been flipped, and then transitioning to neighboring nodes. However, since flipping an edge twice cancels out the operations, the current greedy approach suffices.

Time: O(n) if implemented efficiently by traversing the tree. However, this is probably not necessary in most test cases since the problem can be approached in O(n) time with O(1) space.Space: O(n) in the worst case, to store the DP table.

Trade-offs: Dynamic programming is unnecessary as this greedy approach provides an optimal solution in O(1) space.

✨

Key Insights

1Insight 1: The key insight is that performing the XOR operation twice on the same edge restores the original values. This means we only need to consider performing the operation zero or one time on each edge.
2Insight 2: The parity (even or odd) of the number of nodes XORed with `k` matters. If an even number of nodes have been XORed, the total sum is not affected (as each XOR is paired up). If an odd number of nodes have been XORed with `k`, we should minimize the loss from the XOR operations by XORing the value that results in the least overall difference (absolute difference between the original value and the XORed value).
3Insight 3: The tree structure itself doesn't fundamentally impact the approach other than the XOR operations apply between adjacent nodes. No tree traversal is required

📚

Key Terms Explained

For Students

Array

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

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Tree

Hierarchical structure with nodes. Like a family tree!

Greedy Algorithm

Making the best choice at each step without looking ahead.

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: All nodes have the same value and `k` is chosen such that XORing always decreases the value. The solution correctly identifies this and avoids XORing any nodes if the number of nodes is even, or XORs the node with the minimum loss if the number of nodes is odd.
Edge case 2: `k` is very large. The XOR operation might result in large numbers but this is handled correctly by the `int` or `long` datatype capacity.
Edge case 3: The graph contains only two nodes. The algorithm will still correctly decide whether to XOR or not.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly assuming the tree structure necessitates a more complex graph traversal algorithm. The problem can be solved by only focusing on the node values, and XORing any two adjacent nodes does not depend on visiting every node.

✗Common mistake 2: Not considering the parity of the number of flipped nodes. Failing to account for the situation where an odd number of XOR operations are performed will result in suboptimal solutions. It's necessary to determine which flip operation had the lowest magnitude decrease in sum.

✗Common mistake 3: Not using `long long` in Java/C++ when computing the sum, potentially leading to integer overflow for large datasets.

✅

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: Optimizing network configurations where XOR operations are used for data encryption or error correction. Finding the best configuration that maximizes a certain metric by selectively applying XOR operations.

🏦Real-world scenario 2: Resource allocation where performing a certain operation (analogous to XORing with k) on two linked resources causes both resources to change. The goal is to maximize the total value of the resources after a series of such operations.

💡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

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

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

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: June 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

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

@CodeNinjaJul 14

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

@TechLeadJul 7

❓

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 identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

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 "Find the Maximum Sum of Node Values"?

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 dynamic programming techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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