Main Function: maximumValueSum(nums, k, edges)

Initialise a 2D memoisation array memo with all values set to (-1).
Call the recursive function maxSumOfNodes with:
- Initial index (= 0),
- Initial parity (= 1) (even parity assumed at the start),
- Input array nums,
- XOR value (k), and
- Memoisation array memo.
Return the result of maxSumOfNodes.

Recursive Function: maxSumOfNodes(index, isEven, nums, k, memo)

If all nodes have been processed ((index = nums.length)):
- Return (0) if (isEven = 1), or (INT_MIN) otherwise.
If the result for the current state exists in memo, retrieve and return it.
Explore two possibilities:
- No XOR applied: Calculate the sum by keeping the current value and recursively processing the next node with the same parity.
- XOR applied: Compute the sum by updating the value using (XOR), flipping the parity, and recursively processing the next node.
Memoise the maximum of the two cases for the current state and return it.

Code

[{"language":"Java","code":"class Solution {\n    public long maximumValueSum(int[] nums, int k, int[][] edges) {\n        long[][] memo = new long[nums.length][2];\n        for (long[] row : memo) {\n            Arrays.fill(row, -1);\n        }\n        return calculateMaxSum(0, 1, nums, k, memo);\n    }\n\n    private long calculateMaxSum(int index, int isEven, int[] nums, int k, long[][] memo) {\n        if (index == nums.length) {\n            return isEven == 1 ? 0 : Integer.MIN_VALUE; // Base case: all nodes processed\n        }\n        if (memo[index][isEven] != -1) {\n            return memo[index][isEven]; // Use cached result if available\n        }\n        // Option 1: No XOR operation\n        long noXorSum = nums[index] + calculateMaxSum(index + 1, i

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