# Optimizing Array Values with Bitwise XOR
\n## Intuition\nWhen I first looked at this problem, my goal was to maximize the sum of the array after performing a bitwise XOR operation on its elements with 
\n## Intuition\nWhen I first looked at this problem, my goal was to maximize the sum of the array after performing a bitwise XOR operation on its elements with a given integer `k`. The key observation is that the XOR operation can sometimes increase the value of a number and sometimes decrease it. Therefore, we need to find a way to leverage this operation to get the maximum possible sum of the array.\n\n## Approach\nHere\'s a step-by-step breakdown of my approach:\n\n1. **Calculate the Initial Total Sum**: First, I calculate the total sum of the array elements as a baseline.\n\n2. **Analyze the XOR Differences**: For each element in the array, I compute the difference between the result of the XOR operation and the original value. This difference helps to determine whether the XOR operation is beneficial (i.e., it results in a higher value).\n\n3. **Summing Positive Differences**: I maintain a running total of all positive differences (cases where the XOR operation increases the value). This helps in identifying the potential maximum gain.\n\n4. **Handling Odd Count of Positive Differences**: If the number of positive differences is odd, subtract the smallest absolute difference. This step ensures that the final total remains as high as possible because an odd number of changes might end up decreasing the overall sum when summed together.\n\n5. **Calculate the Final Total**: Add the total positive differences (adjusted for odd count if necessary) to the initial total sum to get the final result.\n\nBy following this method, the solution efficiently determines the maximum possible sum after the XOR operations.\n\n## Complexity\n- **Time complexity**: The solution involves iterating through the array twice: once for calculating the initial sum and once for computing differences and making adjustments. Therefore, the time complexity is $$O(n)$$, where `n` is the number of elements in the array.\n\n- **Space complexity**: The space complexity is $$O(1)$$ because we only use a few extra variables for storing intermediate results, and no additional data structures are required.\n\n\n\n# Code\n```\nclass Solution {\n    public long maximumValueSum(int[] nums, int k, int[][] edges) {\n        long total = 0;\n        for (int num : nums) {\n            total += num;\n        }\n\n        long totalDiff = 0;\n        long diff;\n        int positiveCount = 0;\n        long minAbsDiff = Long.MAX_VALUE;\n        for (int num : nums) {\n            diff = (num ^ k) - num;\n\n            if (diff > 0) {\n                totalDiff += diff;\n                positiveCount++;\n            }\n            minAbsDiff = Math.min(minAbsDiff, Math.abs(diff));\n        }\n        if (positiveCount % 2 == 1) {\n            totalDiff -= minAbsDiff;\n        }\n        return total + totalDiff;\n    }\n}\n```
['Java']