LeetCode 2243. Calculate Digit Sum of a String

You are given a string s consisting of digits and an integer k.

A round can be completed if the length of s is greater than k. In one round, do the following:

1. Divide s into consecutive groups of size k such that the first k characters are in the first group, the next k characters are in the second group, and so on. Note that the size of the last group can be smaller than k.

2. Replace each group of s with a string representing the sum of all its digits. For example, "346" is replaced with "13" because 3 + 4 + 6 = 13.

3. Merge consecutive groups together to form a new string. If the length of the string is greater than k, repeat from step 1.

Return s after all rounds have been completed.

Example 1:

Input: s = "11111222223", k = 3

Output: "135"

Explanation: 

- For the first round, we divide s into groups of size 3: "111", "112", "222", and "23".  

Then we calculate the digit sum of each group: 1 + 1 + 1 = 3, 1 + 1 + 2 = 4, 2 + 2 + 2 = 6, and 2 + 3 = 5.  

 So, s becomes "3" + "4" + "6" + "5" = "3465" after the first round. 

- For the second round, we divide s into "346" and "5".   

Then we calculate the digit sum of each group: 3 + 4 + 6 = 13, 5 = 5.   

So, s becomes "13" + "5" = "135" after second round. Now, s.length <= k, so we return "135" as the answer.

Example 2:

Input: s = "00000000", k = 3

Output: "000"

Explanation: 

We divide s into "000", "000", and "00". 

Then we calculate the digit sum of each group: 

0 + 0 + 0 = 0, 0 + 0 + 0 = 0, and 0 + 0 = 0. s becomes "0" + "0" + "0" = "000", whose length is equal to k, so we return "000".

 需要2个循环，一个是每次字符串的长度>k，另一个是每次去用一个stringbuilder去存储处理的到数据；

下面是代码：

Constraints:

• 1 <= s.length <= 100

• 2 <= k <= 100

• s consists of digits only.

Runtime: 1 ms, faster than 89.94% of Java online submissions for Calculate Digit Sum of a String.

Memory Usage: 40.7 MB, less than 73.27% of Java online submissions for Calculate Digit Sum of a String.