Find the Maximum Sum of Node Values

Updated on 28 May, 2025
Find the Maximum Sum of Node Values header image

Problem Statement

In the context of the problem, you are working with a tree structure consisting of n nodes, each labeled from 0 to n - 1. The tree is defined using a list of edges, where each edge binds two nodes together, creating an undirected connection. Alongside this structural information, each node holds a numerical value specified in the array nums. You are also given an integer k which plays a crucial role in a type of operation you can perform on this tree.

The core task is to maximize the sum of the values of all nodes. You can perform a specific operation to achieve this, which involves choosing any edge, and for the two nodes that the edge connects, you toggle their values using the XOR operation with k. The challenge is to determine the highest possible sum of all node values in the tree after performing this operation any number of times, including the possibility of not performing it at all.

Examples

Example 1

Input:

nums = [1,2,1], k = 3, edges = [[0,1],[0,2]]

Output:

6

Explanation:

Alice can achieve the maximum sum of 6 using a single operation:
- Choose the edge [0,2]. nums[0] and nums[2] become: 1 XOR 3 = 2, and the array nums becomes: [1,2,1] -> [2,2,2].
The total sum of values is 2 + 2 + 2 = 6.
It can be shown that 6 is the maximum achievable sum of values.

Example 2

Input:

nums = [2,3], k = 7, edges = [[0,1]]

Output:

9

Explanation:

Alice can achieve the maximum sum of 9 using a single operation:
- Choose the edge [0,1]. nums[0] becomes: 2 XOR 7 = 5 and nums[1] become: 3 XOR 7 = 4, and the array nums becomes: [2,3] -> [5,4].
The total sum of values is 5 + 4 = 9.
It can be shown that 9 is the maximum achievable sum of values.

Example 3

Input:

nums = [7,7,7,7,7,7], k = 3, edges = [[0,1],[0,2],[0,3],[0,4],[0,5]]

Output:

42

Explanation:

The maximum achievable sum is 42 which can be achieved by Alice performing no operations.

Constraints

  • 2 <= n == nums.length <= 2 * 104
  • 1 <= k <= 109
  • 0 <= nums[i] <= 109
  • edges.length == n - 1
  • edges[i].length == 2
  • 0 <= edges[i][0], edges[i][1] <= n - 1
  • The input is generated such that edges represent a valid tree.

Approach and Intuition

To grasp the solution approach towards achieving the maximum possible sum of the nodes' values, let's break down what happens during the operation and how this influences the sum:

  1. Understand XOR Operation: The XOR operation introduces a toggle effect. If num XOR k results in a higher value than num, then performing the XOR operation is advantageous in increasing the summed value of the tree's nodes.

  2. Observe Patterns and Implications:

    • If toggling a node’s value using the XOR operation yields a higher individual value, you would ideally want to toggle such a node.
    • However, the operation must be performed on pairs of nodes that share an edge. This implies that while you may increase one node's value, you might simultaneously decrease the connected node's value.
  3. Strategy Extraction from Examples:

    • Example 1: The optimal approach involved finding the right edge where the XOR operation maximizes the values of both connected nodes.
    • Example 2: A direct application of the XOR operation on the only available edge provided the result.
    • Example 3: It highlighted a scenario where performing no operations yields the maximum sum, demonstrating the necessity of evaluating the initial configuration before attempting operations.
  4. Optimal Strategy:

    • Begin by evaluating the initial sum of all node values.
    • For each edge, simulate the effect of applying the XOR operation to see if it provides a better sum.
    • Keep track of the highest sum observed.

This problem challenges you to apply both the breadth of algorithmic knowledge, regarding tree traversal and manipulation, and the depth of understanding bitwise operations and their effects on data. The solution would likely need to efficiently traverse the tree while keeping a keen eye on the potential value changes introduced by operations on any edge. The constraint of performing operations any number of times implies that some nodes could be toggled multiple times, either directly or indirectly, through different paths and sequences of operations.

Solutions

  • C++
  • Java
  • Python
cpp
class Solution {
public:
    long long maxValSum(vector<int>& vals, int xor_key, vector<vector<int>>& connections) {
        long long totalSum = 0;
        int evenPosChanges = 0, smallestPosDiff = INT_MAX, largestNegDiff = INT_MIN;
        
        for (int val : vals) {
            int newVal = val ^ xor_key;
            totalSum += val;
            int delta = newVal - val;
            
            if (delta > 0) {
                smallestPosDiff = min(smallestPosDiff, delta);
                totalSum += delta;
                evenPosChanges++;
            } else {
                largestNegDiff = max(largestNegDiff, delta);
            }
        }
        
        if (evenPosChanges % 2 == 0)
            return totalSum;
        
        return max(totalSum - smallestPosDiff, totalSum + largestNegDiff);
    }
};

This solution in C++ provides a method to find the maximum sum of node values in a graph where nodes are represented by their values in a vector, and the edges are represented by another vector (of vectors) indicating connections, which is not used in the solving process. Here's a breakdown of the solution's main steps and logic:

  1. Initialize totalSum to accumulate the sum of original node values.
  2. Use evenPosChanges to track how many times the modified node value increased in comparison with the original node value.
  3. Initialize smallestPosDiff and largestNegDiff to track the smallest positive difference and the largest negative difference, respectively.

Each node value is modified with an XOR operation against a given xor_key, and these steps are followed:

  1. Iterate over each node value:
    • Compute the new value newVal by applying XOR with xor_key.
    • Compute the difference delta between the new value and the original value.
    • Update totalSum directly if this difference is positive and track it by updating smallestPosDiff.
    • If delta is negative, just update largestNegDiff.
  2. After iterating over all nodes, check the number of positive modifications:
    • If evenPosChanges is even, return the totalSum as the answer because no further modifications need balance.
    • If it's odd, adjust totalSum using either smallestPosDiff or largestNegDiff based on which one maximizes the sum.

The answer is derived either directly from the sum of the modified values or adjusted to ensure the sum is maximized when considering necessary corrections for balance. This approach efficiently combines the properties of XOR manipulation with simple conditional checks and arithmetic operations to resolve the problem.

java
class Solution {
    public long calculateMaxSum(int[] values, int modifier, int[][] connections) {
        long accumulatedSum = 0;
        int numPositiveChanges = 0, minPositiveDiff = Integer.MAX_VALUE, maxNegativeDiff = Integer.MIN_VALUE;

        for (int value : values) {
            int adjustedValue = value ^ modifier;
            accumulatedSum += value;
            int changeDifference = adjustedValue - value;
            if (changeDifference > 0) {
                minPositiveDiff = Math.min(minPositiveDiff, changeDifference);
                accumulatedSum += changeDifference;
                numPositiveChanges++;
            } else {
                maxNegativeDiff = Math.max(maxNegativeDiff, changeDifference);
            }
        }

        if (numPositiveChanges % 2 == 0) {
            return accumulatedSum;
        }

        return Math.max(accumulatedSum - minPositiveDiff, accumulatedSum + maxNegativeDiff);
    }
}

You will calculate the maximum sum of node values from an array, considering a specific modifier and connections.

Start by defining a calculateMaxSum method in your Java program that takes an array of integers (values), a modifier integer, and a matrix of integer arrays (connections). This method will return the maximum possible accumulated sum of node values after adjustments with bitwise operations and mathematical calculations.

  • Follow these principles in your code to achieve the required result:

    • Initialize accumulatedSum to keep track of the total value before and after adjustments.
    • Use numPositiveChanges to count how many times a node value increases after adjustment.
    • minPositiveDiff stores the smallest positive change, and maxNegativeDiff stores the largest negative change between the original and adjusted values.
  • For each value in the values array:

    1. Apply a bitwise XOR operation between the value and the modifier to get adjustedValue.
    2. Add the original value to accumulatedSum.
    3. Calculate the difference (changeDifference) between adjustedValue and the original value.
    4. If changeDifference is positive, update minPositiveDiff and additional adjustments to accumulatedSum. Increase the numPositiveChanges counter.
    5. If changeDifference is negative, update maxNegativeDiff with the maximum decrease.
  • Post-process the results:

    • If the count of positive changes is even, the accumulated sum is maximized as it is.
    • If the count is odd, compare and return the greater value between reducing the minimum positive difference from the accumulatedSum or adding the maximum negative difference to it.

This method efficiently consolidates the node values factoring in potential adjustments dictated by the modifier and provides the maximum possible sum under given conditions.

python
class Solution:
    def findMaxValue(self, elements: List[int], offset: int, connections: List[List[int]]) -> int:
        totalSum = 0
        positiveCount = 0
        smallestPositive = 1 << 30
        largestNegative = -1 * (1 << 30)

        for value in elements:
            transformedValue = value ^ offset
            totalSum += value
            delta = transformedValue - value
            if delta > 0:
                smallestPositive = min(smallestPositive, delta)
                totalSum += delta
                positiveCount += 1
            else:
                largestNegative = max(largestNegative, delta)

        # Return even positive deltas' total sum.
        if positiveCount % 2 == 0:
            return totalSum

        # Return max total considering adjustments.
        return max(totalSum - smallestPositive, totalSum + largestNegative)

The provided Python code defines a method findMaxValue to find the maximum sum of node values based on input elements, an offset, and connections.

Exploring Logic:

  • The method initializes the totalSum of all node values, and variables to track the smallest positive delta (smallestPositive) and the largest negative delta (largestNegative).
  • Iterates through each element in the elements list, computes transformedValue as the XOR of the element with the offset, and includes the original value in totalSum.
  • Computes the delta as the difference between transformedValue and the original value.
    • If the delta is positive, it updates the smallestPositive and the positive count while adding the delta to totalSum.
    • If the delta is negative, updates the largestNegative.
  • After processing all elements:
    • If the count of positive deltas is even, it simply returns totalSum.
    • If odd, it determines the maximum sum either by subtracting the smallest positive or adding the largest negative to the totalSum.

Application Insights:

  • The code handles adjustments based on whether the transformation increases or decreases the node value and carefully ensures the maximum sum is achieved under defined constraints.
  • This method's versatility in adapting to differences generated by XOR operations and its cautious approach to edge cases can be robust for scenarios involving data transformation and optimization.

This outline helps understand the core functionality of calculating a potentially maximized sum by considering altered values in terms of bitwise operations and balancing those modifications against an overall measurement of change (either positive or negative).

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