2243. Calculate Digit Sum of a String

2243. Calculate Digit Sum of a String

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
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:

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.
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.
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.

Difficulty : Easy

Solution

Recursive

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
class Solution {
public String digitSum(String s, int k) {
int n = s.length();
if (n <= k) {
return s;
}

StringBuilder ans = new StringBuilder();
int sum = 0;
for (int i = 1; i <= n; i++) {
sum += s.charAt(i - 1) - '0';
if (i % k == 0 || i == n) {
ans.append(sum);
sum = 0;
}
}
return digitSum(ans.toString(), k);
}
}