C# Solution for Find the Original Array of the Prefix XOR Problem
Intuition\n Describe your first thoughts on how to solve this problem. \nThe intuition behind the given solution lies in the property of the XOR operation and h
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe intuition behind the given solution lies in the property of the XOR operation and how it can be used to reconstruct the original array by performing XOR operations in reverse order on the \u2018pref\u2019 array.\n\nThe XOR operation is reversible. Given two numbers A and B, if we perform an XOR operation between them (A ^ B), the result can be used to recover either A or B by performing another XOR operation with the other number (A ^ (A ^ B) = B and B ^ (A ^ B) = A). Using this property, the solution works backward, gradually reconstructing the \u2018arr\u2019 array by reversing the XOR operations carried out on \u2018pref\u2019.     \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1.\tIterative XOR Operations in Reverse: The solution iterates through the \u2018pref\u2019 array in reverse (from the last element towards the first).\n2.\tXOR Operation: At each step, it computes the XOR of \u2018pref[i]\u2019 and \u2018pref[i-1]\u2019 and stores the result in \u2018pref[i]\u2019. This process continues until the first element is reached.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nThe time complexity of this solution is O(n), where \u2018n\u2019 is the number of elements in the \u2018pref\u2019 array. The algorithm processes each element in the array only once, performing XOR operations, which takes constant time.\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nThe space complexity is O(1) as the algorithm operates in place without utilizing any extra space in proportion to the input size. The operations are performed directly on the \u2018pref\u2019 array without using any additional data structures.\n\n# Code\n```\npublic class Solution {\n    public int[] FindArray(int[] pref) {\n        int n = pref.Length;\n\n        for (int i = n - 1; i > 0; i--) {\n            pref[i] = pref[i] ^ pref[i - 1];\n        }\n\n        return pref;\n    }\n}\n```
['C#']