Fcuk strrev
Intuition\n Describe your first thoughts on how to solve this problem. \nstrlen(string_name)\treturns the length of string name.\n2)\tstrcpy(destination, source
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nstrlen(string_name)\treturns the length of string name.\n2)\tstrcpy(destination, source)\tcopies the contents of source string to destination string.\n3)\tstrcat(first_string, second_string)\tconcats or joins first string with second string. The result of the string is stored in first string.\n4)\tstrcmp(first_string, second_string)\tcompares the first string with second string. If both strings are same, it returns 0.\n5)\tstrrev(string)\treturns reverse string.\n6)\tstrlwr(string)\treturns string characters in lowercase.\n![pexels-aleksandar-pasaric-3629227.jpg](https://assets.leetcode.com/users/images/a7ee3882-d120-4642-9acf-45dad42756ac_1696751561.2471952.jpeg)\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nvoid stringev(char *str1)\n{  \n    // declare variable  \n    int i, len, temp;  \n    len = strlen(str1); // use strlen() to get the length of str string  \n      \n    // use for loop to iterate the string   \n    for (i = 0; i < len/2; i++)  \n    {  \n        // temp variable use to temporary hold the string  \n        temp = str1[i];  \n        str1[i] = str1[len - i - 1];  \n        str1[len - i - 1] = temp;  \n    } \n    \n}\n    \nint maximumNumberOfStringPairs(char ** words, int wordsSize){\n    int ho=0;\n \n    for(int i=0;i<wordsSize;i++){\n        stringev(words[i]);\n        for(int j=i+1;j<wordsSize;j++)\n            if(!strcmp(words[i],words[j])){\n            ho++;\n            }\n}\nreturn ho;\n}\n\n\n\n```
['C']