Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.
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.Problem summary: 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.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Dynamic Programming · Greedy · Bit Manipulation · Tree
[1,2,1] 3 [[0,1],[0,2]]
[2,3] 7 [[0,1]]
[7,7,7,7,7,7] 3 [[0,1],[0,2],[0,3],[0,4],[0,5]]
maximum-score-after-applying-operations-on-a-tree)find-number-of-coins-to-place-in-tree-nodes)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #3068: Find the Maximum Sum of Node Values
class Solution {
public long maximumValueSum(int[] nums, int k, int[][] edges) {
long f0 = 0, f1 = -0x3f3f3f3f;
for (int x : nums) {
long tmp = f0;
f0 = Math.max(f0 + x, f1 + (x ^ k));
f1 = Math.max(f1 + x, tmp + (x ^ k));
}
return f0;
}
}
// Accepted solution for LeetCode #3068: Find the Maximum Sum of Node Values
func maximumValueSum(nums []int, k int, edges [][]int) int64 {
f0, f1 := 0, -0x3f3f3f3f
for _, x := range nums {
f0, f1 = max(f0+x, f1+(x^k)), max(f1+x, f0+(x^k))
}
return int64(f0)
}
# Accepted solution for LeetCode #3068: Find the Maximum Sum of Node Values
class Solution:
def maximumValueSum(self, nums: List[int], k: int, edges: List[List[int]]) -> int:
f0, f1 = 0, -inf
for x in nums:
f0, f1 = max(f0 + x, f1 + (x ^ k)), max(f1 + x, f0 + (x ^ k))
return f0
// Accepted solution for LeetCode #3068: Find the Maximum Sum of Node Values
impl Solution {
pub fn maximum_value_sum(nums: Vec<i32>, k: i32, edges: Vec<Vec<i32>>) -> i64 {
let mut f0: i64 = 0;
let mut f1: i64 = i64::MIN;
for &x in &nums {
let tmp = f0;
f0 = std::cmp::max(f0 + x as i64, f1 + (x ^ k) as i64);
f1 = std::cmp::max(f1 + x as i64, tmp + (x ^ k) as i64);
}
f0
}
}
// Accepted solution for LeetCode #3068: Find the Maximum Sum of Node Values
function maximumValueSum(nums: number[], k: number, edges: number[][]): number {
let [f0, f1] = [0, -Infinity];
for (const x of nums) {
[f0, f1] = [Math.max(f0 + x, f1 + (x ^ k)), Math.max(f1 + x, f0 + (x ^ k))];
}
return f0;
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.
Wrong move: Locally optimal choices may fail globally.
Usually fails on: Counterexamples appear on crafted input orderings.
Fix: Verify with exchange argument or monotonic objective before committing.
Wrong move: Recursive traversal assumes children always exist.
Usually fails on: Leaf nodes throw errors or create wrong depth/path values.
Fix: Handle null/base cases before recursive transitions.