Python suboptimal solution
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe key intuition is that the value that minimizes the sum of absolute differences to a set of numbers is the median of that set. Since we are looking for a palindromic number to minimize the cost, we should start our search near the median of the input array nums.\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. **Calculate the Median:** Find the median of the input array `nums`. This serves as the starting point for our search.\n2. **Generate Palindrome Candidates:** Create a few palindrome candidates around the median. This includes the median itself (converted to a palindrome), the immediately smaller palindrome, and the immediately larger palindrome.\n\n    * Mirroring the First Half: The most straightforward way to create a palindrome candidate from a number is to take its first half and append the reverse of that first half (or the first half minus the middle digit if it\'s of odd length).\n    * Finding the Next/Previous Palindrome: Incrementing or decrementing from a given number and checking if it\'s a palindrome is done because sometimes the mirrored candidate isn\'t optimal (it\'s possible the next or previous palindrome is closer to the median and results in a lower cost). Iterating one-by-one to find the next closest palindrome above and below the median guarantees we check the optimal candidates.\n\n3. **Calculate Cost:**  For each palindrome candidate:\n    * Initialize the cost to 0.\n    * Iterate through the input array `nums` and calculate the absolute difference between each element and the palindrome candidate. Add this difference to the cost.\n4. **Minimize Cost:**  Keep track of the minimum cost found across all palindrome candidates.\n# Complexity\n- Time complexity: $$O(n \\log n + k \\cdot n)$$\n    - Calculating the median has a $$O(n \\log n)$$ time complexity in general. Note: In Python `statistics.median` can be optimized to O(n) in some cases, but I\'ll list the more general complexity of computing a median.\n    - Generating palindrome candidates takes O(1) time as we\'re generating at most three candidates no matter how large n is.\n    - Calculating the cost for each candidate takes $$O(n)$$ time. Because there are at most 3 palindrome candidates for each median, we do at most `3 * n` work in this step, so this simplifies to $$O(n)$$.\n\n- Space complexity: $$O(1)$$\n    - The space used is constant regardless of the size of `nums`. We store the median, a few candidate palindromes, and the current cost, all of which require a fixed amount of space. We don\'t store data structures that scale with the input size.\n# Code\n```python3 []\nimport statistics\n\nclass Solution:\n    def minimumCost(self, nums: List[int]) -> int:\n        def is_palindrome(n):\n            return str(n) == str(n)[::-1]\n\n        def generate_palindromes(num):\n            s_num = str(num)\n            n = len(s_num)\n            candidates = []\n            candidates.append(int(s_num[:(n + 1) // 2] + s_num[:n // 2][::-1]))  # First half mirrored\n            if n > 1:\n                prev_num = num - 1\n                while not is_palindrome(prev_num) and prev_num > 0:\n                    prev_num -= 1\n                if prev_num > 0:\n                    candidates.append(prev_num)\n                next_num = num + 1\n                while not is_palindrome(next_num) and next_num < 1000000000:\n                    next_num += 1\n                if next_num < 1000000000:\n                    candidates.append(next_num)\n            return candidates\n\n        median = int(statistics.median(nums))\n        \n        min_cost = float(\'inf\')\n        for palindrome_candidate in generate_palindromes(median):\n            \n            current_cost = 0\n            for x in nums:\n                current_cost += abs(x-palindrome_candidate)\n            min_cost = min(current_cost, min_cost)\n        return min_cost\n```
['Array', 'Math', 'Binary Search', 'Greedy', 'Sorting', 'Python3']