1+ # A string s can be partitioned into groups of size k using the following procedure:
2+ 3+ # The first group consists of the first k characters of the string, the second group
4+ # consists of the next k characters of the string, and so on. Each element can be
5+ # a part of exactly one group.
6+ # For the last group, if the string does not have k characters remaining, a character
7+ # fill is used to complete the group.
8+ # Note that the partition is done so that after removing the fill character from the
9+ # last group (if it exists) and concatenating all the groups in order, the resultant
10+ # string should be s.
11+ 12+ # Given the string s, the size of each group k and the character fill, return a string
13+ # array denoting the composition of every group s has been divided into, using the above procedure.
14+ 15+ 16+ # Example 1:
17+ # Input: s = "abcdefghi", k = 3, fill = "x"
18+ # Output: ["abc","def","ghi"]
19+ # Explanation:
20+ # The first 3 characters "abc" form the first group.
21+ # The next 3 characters "def" form the second group.
22+ # The last 3 characters "ghi" form the third group.
23+ # Since all groups can be completely filled by characters from the string, we do not need to use fill.
24+ # Thus, the groups formed are "abc", "def", and "ghi".
25+ 26+ # Example 2:
27+ # Input: s = "abcdefghij", k = 3, fill = "x"
28+ # Output: ["abc","def","ghi","jxx"]
29+ # Explanation:
30+ # Similar to the previous example, we are forming the first three groups "abc", "def", and "ghi".
31+ # For the last group, we can only use the character 'j' from the string. To complete this group, we add 'x' twice.
32+ # Thus, the 4 groups formed are "abc", "def", "ghi", and "jxx".
33+ 34+ 35+ # Constraints:
36+ 37+ # 1 <= s.length <= 100
38+ # s consists of lowercase English letters only.
39+ # 1 <= k <= 100
40+ # fill is a lowercase English letter.
41+ 42+ 43+ from typing import List
44+ 45+ class Solution :
46+ def divideString (self , s : str , k : int , fill : str ) -> List [str ]:
47+ length = len (s )// k
48+ rmdr = len (s )% k
49+ ans = []
50+ 51+ for i in range (length ):
52+ ans .append (s [i * k :k * (i + 1 )])
53+ if rmdr != 0 :
54+ ans .append (s [- rmdr :].ljust (k , fill ))
55+ return ans
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