Introduction

In this blog post, we will explore how to solve the LeetCode problem "3068" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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] = [ui, vi] indicates that there is an edge between nodes ui and vi 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 following operation any number of times (including zero) 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 Return the maximum possible sum of the values Alice can achieve by performing the operation any number of times. 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.

Explanation

Here's the solution to the problem, with a focus on efficiency and clarity:

High-Level Approach:

• Parity Matters: The core idea is to recognize that applying the XOR operation on an edge twice returns the values to their original state. Therefore, the number of times an edge is used is either 0 or 1, making parity (even or odd) the crucial factor.
• DFS to Determine Edge Usage Parity: We can use Depth-First Search (DFS) to traverse the tree and determine how many times each node's value will be XORed with k. The root node's value is initially XORed 0 or 1 times. The XOR count propagates down the tree.
• Optimization Based on Parity: If the total number of nodes that will be XORed with k is even, then the optimal strategy is to XOR those nodes, as the even count can be considered as a combination of XORs of 0 or 2, resulting in no changes. If the count is odd, it means we must XOR an odd number of nodes with k. In that case, to maximize the sum, we should XOR k with the node that incurs the smallest reduction in value (i.e., the node where abs(nums[i] - (nums[i] ^ k)) is the smallest).

Complexity:

• Runtime: O(n), due to the DFS traversal.
• Space: O(n), for the adjacency list representation of the tree and the recursion stack in DFS (in the worst case, where the tree is a linked list).

Code

    def max_sum(nums, k, edges):
    n = len(nums)
    adj = [[] for _ in range(n)]
    for u, v in edges:
        adj[u].append(v)
        adj[v].append(u)

    xor_count = 0

    def dfs(node, parent, xor_needed):
        nonlocal xor_count
        if xor_needed:
            xor_count += 1

        for neighbor in adj[node]:
            if neighbor != parent:
                dfs(neighbor, node, not xor_needed)

    dfs(0, -1, False)  # Start DFS from node 0, no initial XOR

    total_sum = sum(nums)

    if xor_count % 2 == 0:
        return total_sum
    else:
        min_diff = float('inf')
        for i in range(n):
            diff = abs(nums[i] - (nums[i] ^ k))
            min_diff = min(min_diff, diff)

        return total_sum - min_diff