Easily understandable C++ solution
Intuition\nJust try to make the (i)th and (l-1-i)th characters same with the given condition (The lower value charater shoulb be taken) #where l is the length o
# Intuition\nJust try to make the (i)th and (l-1-i)th characters same with the given condition (The lower value charater shoulb be taken) #where l is the length of th string\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nUse a for loop and iterate upto the middle and try to make the characters same with the given condition\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(n)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n    string makeSmallestPalindrome(string s) {\n        int l=s.length();\n        for(int i=0;i<l/2;i++){\n            if(s[i]!=s[l-1-i]){\n                if(s[i]<s[l-1-i]){ //checking the ascii values of the \n                                   // characters then making them same\n                    s[l-1-i]=s[i];\n                }\n                else{\n                    s[i]=s[l-1-i];\n                }\n            }\n        }\n        return s;\n    }\n};\n```
['String', 'C++']