3068. Find the Maximum Sum of Node Values

HardGreedyBit ManipulationTreeArrayDynamic ProgrammingSorting

Problem Description

There exists an undirected tree with n nodes numbered from 0 to n - 1. You are provided with a 0-indexed 2D integer array edges of length n - 1, where edges[i] = [ui, vi] indicates that there is an edge between nodes ui and vi in the tree. Additionally, you are 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 to maximize the sum of the values of the tree nodes. To achieve this, Alice can perform the following operation any number of times (including none) on the tree:

Choose any edge [u, v] connecting the nodes u and v, and update their values as follows:
- nums[u] = nums[u] XOR k
- nums[v] = nums[v] XOR k

The task is to return the maximum possible sum of the values Alice can achieve by performing the operation any number of times.

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Intuition

The problem essentially revolves around maximizing the sum of an array (derived from nodes of a tree) by using a specific operation involving XOR with a fixed value k. The operation can be applied any number of times on any pair of connected nodes. Here's the reasoning process to arrive at the solution:

XOR Characteristics: When a number is XORed with itself an even number of times, it returns to its original value. Thus, if you can perform an operation on all edges in a path and return to the starting value, it implies no net change, except at the nodes at the ends of the path.
Operation Impact: Given that applying the XOR operation on all path nodes except the start and end nodes results in no change, it implies the question boils down to choosing an optimal subset from the nums array to apply XOR with k.
Subsets and Dynamic Programming: The key is recognizing that we need to choose an even number of elements that will be XORed with k while ensuring the rest remain the same. We can use dynamic programming to maintain two states: f_0 and f_1. Here:
- f_0 keeps track of the maximum sum when an even number of elements have been XORed with k.
- f_1 maintains the maximum sum when an odd number of elements have undergone the XOR operation.
State Transition: As each node value can either be XORed or not in any sequence through the tree, the state transitions are designed to check what gives a better sum at each state:
- If the current element x is added without any operation, update as f_0 + x.
- If the XOR operation is applied, then update as f_1 + (x XOR k) for even numbers and vice versa for odd counts.

Through iterating over the elements in nums and updating these states, we find f_0 as the maximum achievable sum.

Learn more about Greedy, Tree, Dynamic Programming and Sorting patterns.

Solution Approach

To solve this problem efficiently, we use dynamic programming to keep track of two possible states while iterating through the nums array:

Initialization: Start with two variables, f0 and f1, representing the maximum sum when an even or odd number of elements have been XORed with k. Initialize f0 to 0 and f1 to negative infinity (-inf) since initially, no elements have been XORed, and having an odd count of elements XORed is invalid at the start.
Iterative Update: For each element x in nums, update f0 and f1 as follows:
- f0 = max(f0 + x, f1 + (x ^ k)): This transition means, if no more XOR operations turn f0, just add x to f0 to keep even a number of XORs. Alternatively, consider the sum of f1 + (x ^ k) if transitioning from an odd XOR count to an even count is beneficial.
- f1 = max(f1 + x, f0 + (x ^ k)): This allows transitioning f0's sum into an odd count by applying XOR to x, or keeping f1 + x if it maintains a higher odd XOR count.
Final Result: After iterating through the array, f0 holds the maximum sum when the operation has been applied an even number of times, which is the desired value to maximize.

Here's the solution code implementing this approach:

class Solution:
    def maximumValueSum(self, nums: List[int], k: int, edges: List[List[int]]) -> int:
        f0, f1 = 0, float('-inf')  # Initial state
        for x in nums:
            f0, f1 = max(f0 + x, f1 + (x ^ k)), max(f1 + x, f0 + (x ^ k))
        return f0

This approach efficiently manages the sums using two state representations, ensuring a time complexity of O(n) as the algorithm traverses the list of nodes once.

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Example Walkthrough

Let's walk through a small example to illustrate the solution approach:

Consider a tree with 4 nodes (0 to 3) and the following edges:

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

The initial values of the nodes are given by:

nums = [2, 4, 3, 5]

Let's say k = 3.

Our goal is to find the maximum possible sum of the tree nodes' values by performing the operation of XORing values with k on any edge.

Initialization:
- We start with f0 = 0 (max sum with an even number of XORs) and f1 = -inf (max sum with an odd number of XORs).
Iterate Through the Nodes:
- Node 0: Current value is 2.
  - Update f0: f0 = max(f0 + 2, f1 + (2 ^ 3)) = max(0 + 2, -inf) = 2
  - Update f1: f1 = max(f1 + 2, f0 + (2 ^ 3)) = max(-inf + 2, 0 + 1) = 1
- Node 1: Current value is 4.
  - Update f0: f0 = max(f0 + 4, f1 + (4 ^ 3)) = max(2 + 4, 1 + 7) = 8
  - Update f1: f1 = max(f1 + 4, f0 + (4 ^ 3)) = max(1 + 4, 2 + 7) = 9
- Node 2: Current value is 3.
  - Update f0: f0 = max(f0 + 3, f1 + (3 ^ 3)) = max(8 + 3, 9 + 0) = 11
  - Update f1: f1 = max(f1 + 3, f0 + (3 ^ 3)) = max(9 + 3, 8 + 0) = 12
- Node 3: Current value is 5.
  - Update f0: f0 = max(f0 + 5, f1 + (5 ^ 3)) = max(11 + 5, 12 + 6) = 18
  - Update f1: f1 = max(f1 + 5, f0 + (5 ^ 3)) = max(12 + 5, 11 + 6) = 17
Final Result:
- After iterating through all nodes, the maximum sum we can achieve with an even number of XOR applications is stored in f0, which is 18.

This example demonstrates the strategy of using dynamic programming to track the maximum sums with even or odd applications of the XOR operation, ensuring an efficient computation.

Solution Implementation

1from typing import List
2from math import inf  # Importing inf to represent infinity
3
4class Solution:
5    def maximumValueSum(self, nums: List[int], k: int, edges: List[List[int]]) -> int:
6        # Initialize two states:
7        # f0: Maximum sum when not using the operation
8        # f1: Maximum sum when having used the operation at least once
9        f0, f1 = 0, -inf
10
11        # Iterate through each number in the list
12        for x in nums:
13            # Compute the new f0 and f1 considering the current number x:
14            # - f0 can be updated to the maximum between adding x to the previous f0 (no operation),
15            #   and adding (x XOR k) to the previous f1 (using operation)
16            # - f1 can be updated to the maximum between adding x to the previous f1,
17            #   and adding (x XOR k) to the previous f0 (using operation)
18            f0, f1 = max(f0 + x, f1 + (x ^ k)), max(f1 + x, f0 + (x ^ k))
19      
20        # The result is stored in f0, representing the maximum sum achievable
21        # without ending on an operation-used path
22        return f0
23

1class Solution {
2    public long maximumValueSum(int[] nums, int k, int[][] edges) {
3        // Initialize two variables to track maximum sums
4        long withoutXor = 0;          // Represents sum without using XOR
5        long withXor = Long.MIN_VALUE; // Represents sum using XOR at least once
6
7        for (int num : nums) {
8            long temp = withoutXor;
9
10            // Update the sums for this iteration
11            withoutXor = Math.max(withoutXor + num, withXor + (num ^ k));   // Max without using XOR
12            withXor = Math.max(withXor + num, temp + (num ^ k));            // Max using XOR at least once
13        }
14
15        // Return the maximum sum possible
16        return withoutXor;
17    }
18}
19

1class Solution {
2public:
3    // Method to calculate the maximum value sum in an array given XOR operation condition
4    long long maximumValueSum(vector<int>& nums, int k, vector<vector<int>>& edges) {
5        // Initialize the variables to track maximum sums
6        long long f0 = 0;  // f0: Max sum without XOR modification
7        long long f1 = -0x3f3f3f3f;  // f1: Max sum with XOR modification, initialized to a very small number
8      
9        // Iterate through each element in the nums vector
10        for (int x : nums) {
11            // Store the current f0 value to use for f1 calculation
12            long long temp = f0;
13          
14            // Update f0: max sum obtained by adding the current number or applying XOR and adding
15            f0 = max(f0 + x, f1 + (x ^ k));
16          
17            // Update f1: max sum obtained by adding the current number or using previously stored f0
18            f1 = max(f1 + x, temp + (x ^ k));
19        }
20      
21        // Return the maximum sum that can be obtained without using the XOR modification
22        return f0;
23    }
24};
25

1// Function to find the maximum sum value by iterating over the array with XOR operations.
2function maximumValueSum(nums: number[], k: number, edges: number[][]): number {
3    // Initialize f0 and f1 to track the maximum sums with specific conditions
4    let f0: number = 0;
5    let f1: number = -Infinity;
6  
7    // Iterate through each element x in the nums array
8    for (const x of nums) {
9        // Update f0 and f1 by considering the maximum sum obtained 
10        // by adding the current element x directly or with an XOR operation with k
11        const newF0 = Math.max(f0 + x, f1 + (x ^ k));
12        const newF1 = Math.max(f1 + x, f0 + (x ^ k));
13        f0 = newF0;
14        f1 = newF1;
15    }
16  
17    // f0 holds the maximum possible sum while adhering to the conditions
18    return f0;
19}
20
21// Define the edges parameter separately, as it's required by the function 
22// signature but is not used in the current implementation.
23

Time and Space Complexity

The time complexity of the code is O(n), where n is the length of the array nums. This is due to the loop which iterates through each element in the nums array exactly once, performing constant time operations.

The space complexity of the code is O(1) because the algorithm uses a fixed amount of additional space (variables f0 and f1) that does not depend on the size of the input.

Learn more about how to find time and space complexity quickly.

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