C# Greedy with Two Pointer Solution
Intuition\n-  Describe your first thoughts on how to solve this problem. First Thoughts: The intuition behind solving this problem is to realize that the array 
# Intuition\n- <!-- Describe your first thoughts on how to solve this problem. -->First Thoughts: The intuition behind solving this problem is to realize that the array you are given (nums) is actually the result of performing an operation that involves an original array with a certain offset (k). The key challenge here is to figure out the original elements and the correct value of k.\n\n- The main observation is that the original array consists of pairs in nums that are separated by 2 * k. Therefore, the approach is to iterate through possible values for k and check if they can successfully reconstruct the original array.\n___\n# Approach\n- <!-- Describe your approach to solving the problem. -->First, sort the given array to make it easier to identify potential pairs.\n- Start by trying different values for k. Since k must be positive, a natural place to look is the difference between the smallest element (nums[0]) and other elements in the sorted list.\n- Use the first two elements (nums[0] and nums[i]) to calculate a candidate value for k. Only valid candidates are considered if k > 0 and the difference is even ((nums[i] - nums[0]) % 2 == 0).\n- Once a candidate k is identified, use a two-pointer approach to verify whether all elements of the given array can be paired in a way that follows the required pattern (nums[right] == nums[left] + 2 * k). Use a visited array to mark which elements have been used.\n- If a candidate value for k successfully pairs all elements, the original array is reconstructed and returned.\n___\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->O(n^2), where \'n\' is the length of the \'nums\' array.\n- The outer loop iterates over the possible values for \'k\', which is O(n).\n- For each candidate \'k\', the two-pointer traversal takes O(n) to find pairs in the original array.\n___\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->O(n), where \'n \'is the length of the \'nums\' array.\n- The \'visited\' boolean array takes O(n) space.\n- The \'originalArray\' list also takes O(n) space.\n___\n\n# Code\n```csharp []\nusing System;\nusing System.Collections.Generic;\nusing System.Linq;\n\npublic class Solution {\n    public int[] RecoverArray(int[] nums) {\n        Array.Sort(nums);\n        int n = nums.Length / 2;\n        \n        for (int i = 1; i < nums.Length; i++) {\n            int potentialK = (nums[i] - nums[0]) / 2;\n            if (potentialK <= 0 || (nums[i] - nums[0]) % 2 != 0) continue;\n            \n            int k = potentialK;\n            var originalArray = new List<int>();\n            var visited = new bool[nums.Length];\n            \n            int left = 0, right = i;\n            while (right < nums.Length) {\n                if (visited[left]) {\n                    left++;\n                    continue;\n                }\n                if (visited[right] || nums[right] != nums[left] + 2 * k) {\n                    right++;\n                    continue;\n                }\n                \n                originalArray.Add(nums[left] + k);\n                visited[left] = true;\n                visited[right] = true;\n                left++;\n                right++;\n            }\n            \n            if (originalArray.Count == n) {\n                return originalArray.ToArray();\n            }\n        }\n        \n        return new int[0];\n    }\n}\n\n//  Sorting the Input: First, we sort nums to ensure we can easily find pairs.\n//  Finding Possible k: We iterate through possible pairs to determine a potential k. This is done by using the difference between the smallest value (nums[0]) and another value (nums[i]). k must be positive, and we ensure it divides evenly.\n//  Pairing Values: We use a greedy approach to pair values. For each k, we attempt to pair each element in nums with another value that matches the expected difference (2 * k). We skip elements that have already been visited.\n//  Checking Validity: If we successfully pair all values and reconstruct the original array, we return it.\n```
['Array', 'Hash Table', 'Two Pointers', 'Sorting', 'C#']