Simple Solution With Explanation inside the code Using different Methods
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n- 29ms\nBeats 57.38%of users with Java\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- 54.46MB\nBeats 95.41%of users with Java\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\npublic class Solution {\n    public long minimumCost(int[] nums) {\n        // Initialize the result to the maximum possible long value\n        long res = Long.MAX_VALUE;\n\n        // Get the length of the input array\n        int size = nums.length;\n\n        // Sort the input array in ascending order\n        Arrays.sort(nums);\n\n        // Find the median of the sorted array\n        int median = findMedian(nums, size);\n\n        // Find the nearest palindrome to the median\n        int palin = NearestNeighbour(median);\n\n        // Initialize a variable to store the previous palindrome\n        int prevPal = palin;\n\n        // Iterate over four possible scenarios for changing the palindrome\n        for (int i = 0; i < 4; i++) {\n            long temp = 0;\n\n            // Calculate the cost for the current palindrome\n            for (int num : nums) {\n                temp += Math.abs(num - palin);\n            }\n\n            // Update the result with the minimum cost\n            res = Math.min(temp, res);\n\n            // Update the palindrome based on the current scenario\n            if (i == 0) {\n                palin = changePalin(palin, -1);\n            } else if (i == 1) {\n                palin = changePalin(prevPal, 1);\n            } else {\n                // Find the single-digit number in the array\n                int s = 0;\n                for (int num : nums) {\n                    if (String.valueOf(num).length() == 1) {\n                        s = num;\n                    }\n                }\n                palin = s;\n            }\n        }\n\n        // Return the minimum cost\n        return res;\n    }\n\n    // Method to change the palindrome based on the given change\n    public int changePalin(int palin, int change) {\n        String t = String.valueOf(palin);\n        char sarr[] = t.toCharArray();\n\n        // Update the left and right parts of the palindrome\n        if (t.length() % 2 == 0) {\n            int l = t.length() / 2 - 1;\n            int r = l + 1;\n            updateDigit(sarr, l, change);\n            updateDigit(sarr, r, change);\n        } else {\n            int l = t.length() / 2;\n            updateDigit(sarr, l, change);\n        }\n\n        // Convert the updated character array to an integer\n        return Integer.valueOf(new String(sarr));\n    }\n\n    // Helper method to update a digit in the character array\n    private void updateDigit(char[] sarr, int index, int change) {\n        if (sarr[index] == \'0\' && change == -1) {\n            sarr[index] = \'9\';\n        } else if (sarr[index] == \'9\' && change == 1) {\n            sarr[index] = \'0\';\n        } else {\n            sarr[index] = (char) (((sarr[index] - \'0\') + change) + \'0\');\n        }\n    }\n\n    // Method to find the nearest palindrome to a given number\n    public int NearestNeighbour(int median) {\n        String s = String.valueOf(median);\n        char sarr[] = s.toCharArray();\n        int size = s.length();\n        int l = 0;\n        int r = 0;\n\n        // Determine the left and right indices based on the length of the string\n        if (size % 2 == 0) {\n            l = (size / 2) - 1;\n            r = l + 1;\n        } else {\n            l = (size / 2) - 1;\n            r = l + 2;\n        }\n\n        // Mirror the left side to the right side to create a palindrome\n        for (int i = r; i < size; i++) {\n            sarr[i] = sarr[l--];\n        }\n\n        // Convert the character array back to an integer\n        return Integer.valueOf(new String(sarr));\n    }\n\n    // Method to find the median of an array\n    public int findMedian(int[] nums, int size) {\n        if (size % 2 == 0) {\n            int x = size / 2;\n            int y = x - 1;\n            return (nums[x] + nums[y]) / 2;\n        } else {\n            return nums[size / 2];\n        }\n    }\n}\n\n```
['Java']