C++ || Easy to Understand
Approach\n1. Sort the Input Array:\n   - The code begins by sorting the input vector nums in ascending order using sort(nums.begin(), nums.end()).\n2. Determine
# Approach\n1. **Sort the Input Array:**\n   - The code begins by sorting the input vector `nums` in ascending order using `sort(nums.begin(), nums.end())`.\n2. **Determine the Middle Element:**\n   - It finds the middle element of the sorted array by accessing `nums[n/2]`, where `n` is the size of the array.\n3. **Find Left and Right Nearest Palindromes:**\n   - It calculates the left and right nearest palindromes for the middle element using the `leftNearestPalindrome` and `rightNearestPalindrome` functions.\n4. **Check Palindrome:**\n   - The `isPalindrome` function checks if a given number is a palindrome by converting it to a string and comparing characters from both ends.\n5. **Calculate Cost Function:**\n   - The `minCost` function calculates the cost of transforming the array elements to a given value `x` by summing the absolute differences.\n6. **Calculate Cost for Left and Right Palindromes:**\n   - It calculates the cost for transforming the array elements to the left and right nearest palindromes.\n7. **Return Minimum Cost:**\n   - The minimum cost between the left and right palindrome transformations is returned as the result.\n8. **Time Complexity:**\n   - The time complexity is O(N log N) due to the sorting operation.\n9. **Space Complexity:**\n   - The space complexity is O(1) as the additional space used is constant.\n10. **Overall Approach:**\n   - The code aims to find the minimum cost of transforming the array elements to either the left or right nearest palindrome of the middle element after sorting. This is done by iterating through possible palindrome values and calculating the cost for each transformation. The final result is the minimum cost among these options.\n\n# Complexity\n- Time complexity:\n$$O(N * Log N)$$\n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\nclass Solution {\nprivate:\n    // Check if a given number is a palindrome\n    bool isPalindrome(int num){\n        // Convert the number to a string for easy comparison of digits\n        string str = to_string(num);\n        int i = 0, j = str.length() - 1;\n\n        // Iterate through the digits from both ends, checking for equality\n        while(i < j){\n            if(str[i] != str[j]) return false;\n            i += 1;\n            j -= 1;\n        }\n\n        // If the loop completes without returning false, the number is a palindrome\n        return true;\n    }\n\n    // Calculate the total cost of transforming array elements to a given value x\n    long long minCost(vector<int>& nums, int x){\n        long long cost = 0;\n        int n = nums.size();\n\n        // Calculate the absolute differences and sum them up\n        for(int i = 0; i < n; i += 1){\n            cost += abs(nums[i] - x);\n        }\n\n        // Return the total cost\n        return cost;\n    }\n\n    // Find the nearest palindrome to the left of a given number\n    int leftNearestPalindrome(int num){\n        // Decrement the number until a palindrome is found\n        while(!isPalindrome(num)){\n            num -= 1;\n        }\n\n        // Return the left nearest palindrome\n        return num;\n    }\n\n    // Find the nearest palindrome to the right of a given number\n    int rightNearestPalindrome(int num){\n        // Increment the number until a palindrome is found\n        while(!isPalindrome(num)){\n            num += 1;\n        }\n\n        // Return the right nearest palindrome\n        return num;\n    }\n\npublic:\n    // Find the minimum cost of transforming array elements to a palindrome\n    long long minimumCost(vector<int>& nums) {\n        // Sort the array in ascending order\n        sort(nums.begin(), nums.end());\n        int n = nums.size();\n        // Find the middle element of the sorted array\n        int mid = nums[n/2];\n\n        // Find the left and right nearest palindromes for the middle element\n        int left = leftNearestPalindrome(mid);\n        int right = rightNearestPalindrome(mid);\n\n        // Return the minimum cost between transforming to left and right nearest palindromes\n        return min(minCost(nums, left), minCost(nums, right));\n    }\n};\n\n```
['Math', 'Sorting', 'C++']