Class Solution
- java.lang.Object
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- g2101_2200.s2138_divide_a_string_into_groups_of_size_k.Solution
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public class Solution extends Object
2138 - Divide a String Into Groups of Size k.Easy
A string
scan be partitioned into groups of sizekusing the following procedure:- The first group consists of the first
kcharacters of the string, the second group consists of the nextkcharacters of the string, and so on. Each character can be a part of exactly one group. - For the last group, if the string does not have
kcharacters remaining, a characterfillis used to complete the group.
Note that the partition is done so that after removing the
fillcharacter from the last group (if it exists) and concatenating all the groups in order, the resultant string should bes.Given the string
s, the size of each groupkand the characterfill, return a string array denoting the composition of every groupshas been divided into, using the above procedure.Example 1:
Input: s = “abcdefghi”, k = 3, fill = “x”
Output: [“abc”,“def”,“ghi”]
Explanation:
The first 3 characters “abc” form the first group.
The next 3 characters “def” form the second group.
The last 3 characters “ghi” form the third group.
Since all groups can be completely filled by characters from the string, we do not need to use fill.
Thus, the groups formed are “abc”, “def”, and “ghi”.
Example 2:
Input: s = “abcdefghij”, k = 3, fill = “x”
Output: [“abc”,“def”,“ghi”,“jxx”]
Explanation:
Similar to the previous example, we are forming the first three groups “abc”, “def”, and “ghi”.
For the last group, we can only use the character ‘j’ from the string.
To complete this group, we add ‘x’ twice. Thus, the 4 groups formed are “abc”, “def”, “ghi”, and “jxx”.
Constraints:
1 <= s.length <= 100sconsists of lowercase English letters only.1 <= k <= 100fillis a lowercase English letter.
- The first group consists of the first
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Constructor Summary
Constructors Constructor Description Solution()
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