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:
[u, v] connecting the nodes u and v, and update their values as follows:
nums[u] = nums[u] XOR knums[v] = nums[v] XOR kReturn 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 * 1041 <= k <= 1090 <= nums[i] <= 109edges.length == n - 1edges[i].length == 20 <= edges[i][0], edges[i][1] <= n - 1edges represent a valid tree.This approach involves checking all possible operations on each edge and calculating the resulting sum by applying the XOR operation. We must continuously update and evaluate for every operation on each pair of nodes linked by an edge to see if the resulting sum after the operation is beneficial.
This naive strategy will consider 2^n operations (since each edge can have 2 XOR states per pair), making it inefficient for larger datasets. Ideal for understanding XOR effects but not recommended for large n.
This C function calculates the maximum possible sum by performing ARC XOR operations optimally on each edge. The initial sum is calculated, then for each edge, the potential new sum after the operation is checked and updated if beneficial.
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Time Complexity: O(n), where n is the number of nodes. Each iteration is constant time.
Space Complexity: O(1), since additional space used is constant.
Instead of brute checking each edge operation individually, observe that XOR operation symmetry allows flipping of node values to be optimal only when advantageous. Analyze and preprocess which nodes when XOR'd with k will increase their value and prioritize these changes globally without redundant individual edge checks.
This idea is crucially about observing which nodes independently contribute more value when toggled and using the tree structure to apply XOR operations optimally.
This C code efficiently determines the maximum sum by comparing the original sum to the sum if every node were XOR'd independently with k. This leverages XOR symmetry to quickly evaluate if any operation is beneficial globally.
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Time Complexity: O(n), as it processes each node individually once.
Space Complexity: O(1), using a fixed amount of extra space.
| Approach | Complexity |
|---|---|
| Brute Force Approach | Time Complexity: O(n), where n is the number of nodes. Each iteration is constant time. |
| Optimized Evaluation | Time Complexity: O(n), as it processes each node individually once. |
Binary Tree Maximum Path Sum - DFS - Leetcode 124 - Python • NeetCode • 187,670 views views
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