# 2138. Divide a String Into Groups of Size k """ a string 's' can be partitioned into groups of size 'k' using the following procedure: the first group consists of the first 'k' characters of the string, the second group consists of the next 'k' characters of the string, ans so on. each element can be a part of exactly one group. for the last group, if the string does not have 'k' characters remaining, a character 'fill' is used to complete the group. note that the partition is done so that after removing the 'fill' character from the last group, and concatenating all the groups in oder, the resultant string should be 's'. given a string 's', the size of each group 'k', and the character 'fill', return a string array denoting the composition of every group 's' has been divided into. """ class Solution(object): def divideString(self, s, k, fill): """ :type s: str :type k: int :type fill: str :rtype: List[str] """ n = len(s) groups, ans = (n + k - 1) // k, [] for i in range(groups): curr = "" for j in range(k): idx = i * k + j if idx < n: curr += s[idx] else: curr += fill ans.append(curr) return ans if __name__ == "__main__": obj = Solution() print(obj.divideString(s="abcdefghi", k=3, fill="x")) print(obj.divideString(s="abcdefghij", k=3, fill="x"))