C++ || Easy || Best Explanation || O(n) time n space
Intuition:\nThe problem is to divide a given string into groups of size k, such that each group is made up of k characters, and any remaining characters are fil
# Intuition:\nThe problem is to divide a given string into groups of size k, such that each group is made up of k characters, and any remaining characters are filled up with a given fill character. To solve this problem, the code first checks if the given string can be divided evenly into k-sized groups. If not, it calculates the number of fill characters needed and adds them to the end of the string. Then, it creates groups of size k by looping over the characters in the string and adding each character to a temporary string until the group size reaches k. The completed group strings are added to a vector, and the loop continues until all characters have been processed.\n\n# Approach:\nThe code takes the following approach to solve the problem:\n\nCreate an empty vector to store the divided strings.\nCheck if the given string can be divided evenly into k-sized groups. If not, calculate the number of fill characters needed and add them to the end of the string.\nLoop over the characters in the string and create groups of size k by adding each character to a temporary string until the group size reaches k.\nAdd the completed group strings to the vector.\nContinue the loop until all characters have been processed.\nReturn the vector containing all the divided strings.\n# Time Complexity:\nThe time complexity of the code is O(n), where n is the length of the input string. This is because the code loops over the characters in the string once to add fill characters (if necessary) and create groups of size k.\n\n# Space Complexity:\nThe space complexity of the code is O(n), where n is the length of the input string. This is because the code creates an empty vector to store the divided strings, and a temporary string to store each group of size k. The size of the vector and the temporary string can be up to n. Additionally, if fill characters are added to the string, the space complexity increases by the number of fill characters added, which can be up to k.\n\n# Code\n```\nclass Solution {\npublic:\n    vector<string> divideString(string s, int k, char fill) {\n        vector<string>v;  // create an empty vector to store the divided strings\n        if(s.size()%k!=0){  // check if the string cannot be divided evenly into k-sized groups\n            int next = ceil(float(s.size())/float(k));  // calculate the number of groups needed\n            int rem = (next*k) - s.size();  // calculate the number of fill characters needed\n            for(int i =0;i<rem;i++){  // add the fill characters to the end of the string\n                s+=fill;\n            }\n        }\n        int temp = 0;  // initialize a counter variable to keep track of the group size\n        string str = "";  // initialize an empty string to store each group\n        for(int i =0;i<s.size();){  // loop over the characters in the string\n            str = "";  // reset the string for each new group\n            while(temp != k){  // loop until the group size reaches k\n               str+=s[i];  // add the current character to the group string\n               i++;  // move to the next character in the string\n               temp++;  // increment the group size counter\n           }\n            v.push_back(str);  // add the completed group string to the vector\n            temp = 0;  // reset the group size counter for the next group\n        }\n        return v;  // return the vector containing all the divided strings\n    }\n};\n\n```
['C++']