## [leetCode] 482. License Key Formatting (Python) 본문

Programming/코딩 1일 1문제

### [leetCode] 482. License Key Formatting (Python)

사용자 솜씨좋은장씨 2020. 10. 31. 19:21

You are given a license key represented as a string S which consists only alphanumeric character and dashes. The string is separated into N+1 groups by N dashes.

Given a number K, we would want to reformat the strings such that each group contains exactly K characters, except for the first group which could be shorter than K, but still must contain at least one character. Furthermore, there must be a dash inserted between two groups and all lowercase letters should be converted to uppercase.

Given a non-empty string S and a number K, format the string according to the rules described above.

Example 1:

``````Input: S = "5F3Z-2e-9-w", K = 4

Output: "5F3Z-2E9W"

Explanation: The string S has been split into two parts, each part has 4 characters.
Note that the two extra dashes are not needed and can be removed.``````

Example 2:

``````Input: S = "2-5g-3-J", K = 2

Output: "2-5G-3J"

Explanation: The string S has been split into three parts, each part has 2 characters except the first part as it could be shorter as mentioned above.``````

Note:

1. The length of string S will not exceed 12,000, and K is a positive integer.
2. String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
3. String S is non-empty.

Solution

``````class Solution:
def licenseKeyFormatting(self, S: str, K: int) -> str:
num_range = list(range(0, 10))
S_String = "".join(S.upper().split("-"))

list_S = list(S_String)
if len(list_S) % K == 0:
for i in range(len(list_S) // K):
elif len(list_S) % K != 0:
cut_num = len(list_S) % K

list_S = list_S[cut_num:]

for i in range(len(list_S) // K):