C# Prefix Sum or Difference Array Solution
Intuition\n Describe your first thoughts on how to solve this problem. Pair Sums: For each pair of numbers (nums[i], nums[n-1-i]), calculate the possible sums.\
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->Pair Sums: For each pair of numbers (nums[i], nums[n-1-i]), calculate the possible sums.\nFrequency and Adjustments: Instead of trying to directly manipulate sums, let\'s focus on the frequency of each sum and how many moves it takes to reach a common sum.\nWe can create a delta array where:\n\nExact matches: We decrease the number of required changes for sums that can be reached exactly by the pair without any modifications.\nOne change needed: We handle scenarios where only one element needs to be changed.\nTwo changes needed: Finally, we handle cases where both elements need to be changed.\n___\n\n# Approach\n<!-- Describe your approach to solving the problem. -->//  Delta Array:\n//  *We start by assuming 2 changes for each possible sum.\n//  *For sums that can be reached with fewer changes, we adjust the delta array accordingly.\n//  *delta[low] -= 1: Indicates that sums greater than or equal to low need one less move.\n//  *delta[sum] -= 1: Subtracts one more move for the exact sum.\n//  *delta[sum + 1] += 1: Indicates that sums greater than the exact sum need one more move.\n//  *delta[high + 1] += 1: Marks that sums beyond high go back to needing 2 changes.\n//  Final Calculation:\n//  *Iterate through all possible sums and calculate the minimum number of moves required.\n___\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->O(n + limit).\n*The loop that processes pairs runs in O(n/2), which is O(n) overall.\n*The final loop that computes the minimum moves from the \'delta\' array runs in O(2 x limit), creating the overall time complexity of O(n + limit).\n___\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->O(limit). The \'delta\' array has a size proportional to 2 X limit, which is the main space usage.\nTherefore, the space complexity is O(limit).\n___\n\n# Code\n```csharp []\npublic class Solution {\n    public int MinMoves(int[] nums, int limit) {\n        int n = nums.Length;\n        int[] delta = new int[2 * limit + 2];\n\n        for (int i = 0; i < n / 2; i++) {\n            int a = nums[i];\n            int b = nums[n - 1 - i];\n            \n            int low = Math.Min(a, b) + 1;\n            int high = Math.Max(a, b) + limit;\n            int sum = a + b;\n\n            delta[2] += 2; // Default: 2 changes required for every sum\n            delta[low] -= 1; // Decrease to 1 change if within this range\n            delta[sum] -= 1; // If we hit the exact sum, subtract 1 more (total 0 changes needed)\n            delta[sum + 1] += 1; // After the exact sum, increment (1 more change required)\n            delta[high + 1] += 1; // After the range, go back to 2 changes\n        }\n\n        int minMoves = int.MaxValue;\n        int currentMoves = 0;\n        for (int i = 2; i <= 2 * limit; i++) {\n            currentMoves += delta[i];\n            minMoves = Math.Min(minMoves, currentMoves);\n        }\n\n        return minMoves;\n    }\n}\n```
['Array', 'Prefix Sum', 'C#']