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Abhishek Chaudhary

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# 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".

Constraints:

• `1 <= s.length <= 100`
• `2 <= k <= 100`
• `s` consists of digits only.

SOLUTION:

``````class Solution:
def digitSum(self, s: str, k: int) -> str:
while len(s) > k:
nexts = ""
for i in range(0, len(s), k):
nexts += str(sum(int(d) for d in s[i:i+k]))
s = nexts
return s
``````