Easy
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:
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
.s
with a string representing the sum of all its digits. For example, "346"
is replaced with "13"
because 3 + 4 + 6 = 13
.k
, repeat from step 1
.Return s
after all rounds have been completed.
Example 1:
Input: s = “11111222223”, k = 3
Output: “135”
Explanation:
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.
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.public class Solution {
public String digitSum(String s, int k) {
while (s.length() > k) {
int n = s.length();
int count = 0;
int sum = 0;
StringBuilder sb = new StringBuilder();
for (int i = 0; i < n; i++) {
if (count == k) {
sb.append(sum);
sum = 0;
count = 0;
}
sum += s.charAt(i) - '0';
count++;
}
sb.append(sum);
s = sb.toString();
}
return s;
}
}