Simple C++ Solution with line-by-line comments
\nclass Solution {\nprivate:\n    // method to check palindrome\n    bool isPalindrome(int givenNum)\n    {\n        long long sum = 0, temp = givenNum;\n      
```\nclass Solution {\nprivate:\n    // method to check palindrome\n    bool isPalindrome(int givenNum)\n    {\n        long long sum = 0, temp = givenNum;\n        \n        while(givenNum > 0){\n            sum = (sum * 10) + (givenNum % 10);\n            givenNum /= 10;\n        }\n        \n        return (sum == temp);\n    }\n    \n    // method to calculate cost, if we modifying all the array elements to palindromicNum \n    long long calCost(vector<int> &nums, int palindromicNum){\n        \n        long long cost = 0;\n        \n        for(const int& num : nums)\n        {\n            cost += abs(palindromicNum - num);\n        }\n        \n        return cost;\n    }\n    \npublic:\n    long long minimumCost(vector<int>& nums) {\n        \n        int n = nums.size();\n        \n        if(n == 1)\n            return 0;\n        \n        // sort the array so that, we can find the middle element \n        sort(nums.begin(), nums.end());\n        \n        /*\n          if n is odd, \n              midEle will be present in middle element of an array i.e n/2 index \n          \n          if n is even,\n              midEle will be present in middle two elements of an array i.e n/2 index + (n+1)/2 index\n        */ \n        int midEle = (n & 1) ? nums[(n/2)] : (nums[(n/2)] + nums[(n+1)/2])/2 ;\n     \n        long long ans = 1e15;\n        \n        // if middle element is palindrome, then calculate cost \n        ans = (isPalindrome(midEle)) ? calCost(nums, midEle) : ans;\n        \n        long long low = midEle - 1, high = midEle + 1;\n        \n        long long left = -1, right = -1;\n        \n        // run the loop until both left and right having some valid element\n        while(left == -1 || right == -1)\n        {\n            // if low is palindrome, it can be our left \n            if(left == -1 && isPalindrome(low))\n            {\n                left = low;\n            }\n            \n            // if high is palindrome, it can be our right\n            if(right == -1 && isPalindrome(high))\n            {\n                right = high;\n            }\n            \n           // else, find all the possibilities of low and high  \n           low--, high++;\n        }\n        \n        // find min cost for both left and right\n        ans = min(ans, min(calCost(nums, left), calCost(nums, right)));\n        \n        return ans;\n    }\n};\n```
['C++']