O(n + m) time easy C++ solution, beat 100%
Intuition\nThe problem involves finding the minimum cost to transform a list of integers into palindromic numbers. The intuition behind solving this problem lie
# Intuition\nThe problem involves finding the minimum cost to transform a list of integers into palindromic numbers. The intuition behind solving this problem lies in understanding that each number in the array can be individually converted into its closest palindromic number. The cost of converting each number is the absolute difference between the original number and its nearest palindromic counterpart. By iterating over the array and calculating these costs, we can determine the minimum cost to make all numbers palindromic. A key aspect to consider is efficiently finding the closest palindromic number for each element in the array.\n\n\n# Approach\n**Check Palindromicity:** Implement a function isPalindromic to check if a number is palindromic. This is done by reversing the number and comparing it with the original.\n**Calculate Individual Costs:** Implement calculateCost to compute the total cost of converting each number in the array to a specific target number.\n**Find Closest Palindromic Number:** Implement closestPalindromic to find the nearest palindromic number to a given number. This is done by incrementing or decrementing the number until a palindromic number is found.\n**Optimize Over All Elements:** In minimumCost, iterate over the range of numbers in the array and calculate the cost of converting all numbers to the closest palindromic number within this range.\n**Consider Edge Cases:** Ensure that negative numbers are handled correctly (they can\'t be palindromic), and optimize the search range based on the array\'s median.\n\n# Complexity\n- Time complexity:\n$$O(n + m)$$: which n is the length of nums and m represent to find the closest palindromic number. \n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\nclass Solution {\npublic:\n    bool isPalindromic(int num) {\n        if (num < 0) return false;\n        int reversed = 0, original = num;\n        while (num != 0) {\n            reversed = reversed * 10 + num % 10;\n            num /= 10;\n        }\n        return original == reversed;\n    }\n\n    long long calculateCost(const vector<int>& nums, int target) {\n        long long cost = 0;\n        for (int num : nums) {\n            cost += abs(num - target);\n        }\n        return cost;\n    }\n    \n    int closestPalindromic(int num, int step) {\n        while (!isPalindromic(num)) {\n            num += step;\n        }\n        return num;\n}\n\n    long long minimumCost(vector<int>& nums) {\n        long long minCost = LLONG_MAX;\n        int maxNum = *max_element(nums.begin(), nums.end());\n        int minNum = *min_element(nums.begin(), nums.end());\n        nth_element(nums.begin(), nums.begin() + nums.size() / 2, nums.end());\n        int medium = nums[nums.size() / 2];\n\n        minCost = min(calculateCost(nums, closestPalindromic(minNum, -1)), minCost);\n        minCost = min(calculateCost(nums, closestPalindromic(maxNum, 1)), minCost);\n        minCost = min(calculateCost(nums, closestPalindromic(medium, -1)), minCost);\n        minCost = min(calculateCost(nums, closestPalindromic(medium, 1)), minCost);\n\n        return minCost;\n    }\n};\n\n```
['C++']