C++ solution beats 100%. exPlained
Intuition\n Describe your first thoughts on how to solve this problem. \n\nMy initial idea is to use a sliding window approach to determine the minimum number o
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nMy initial idea is to use a sliding window approach to determine the minimum number of moves required to make all the elements in the array the same.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\nMy approach is to use a sliding window approach to determine the minimum number of moves required to make all the elements in the array the same. We can start with a window size of two and slide it through the array. If the elements in the window are the same, then we don\'t need to make any changes to the array. If the elements are different, then we can make a move to bring them both to the same value. After making all the moves, we can look at the resulting array and calculate the minimum number of moves required.\n\nTo optimize the approach, we can use a prefix sum array to keep track of the sum of elements in the array up to a certain index. We can also use a diff array to keep track of the difference between the elements in the array. By using a diff array, we can quickly calculate the number of moves required to make the elements in the array the same.\n\n# Complexity\n\n- Time complexity: $$O(n)$$\n- Space complexity: $$O(n)$$\n# Code\n```\n#pragma GCC optimize("Ofast","inline","fast-math","unroll-loops","no-stack-protector")\n#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native","f16c")\nstatic const auto fast = []() {ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0); return 0; } ();\nclass Solution {\npublic:\n    int minMoves(vector<int>& nums, int limit) {\n        int n = nums.size();\n        vector<int> diff(limit * 2 + 2);\n        for (int i = 0; i < n / 2; ++i) {\n            int a = nums[i], b = nums[n - i - 1];\n            diff[2] += 2;\n            diff[min(a, b) + 1] -= 1;\n            diff[a + b] -= 1;\n            diff[a + b + 1] += 1;\n            diff[max(a, b) + limit + 1] += 1;\n        }\n        int ans = n, cur = 0;\n        for (int i = 2; i <= limit * 2; ++i) {\n            cur += diff[i];\n            ans = min(ans, cur);\n        }\n        return ans;\n    }\n};\n```
['Array', 'Hash Table', 'Prefix Sum', 'C++']