Formatted question description: https://leetcode.ca/all/482.html
482. License Key Formatting
Level
Easy
Description
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:
- The length of string S will not exceed 12,000, and K is a positive integer.
- String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
- String S is non-empty.
Solution
First turn all the letters in S to upper case and remove all dashes from S. Then loop over S from right to left and add a dash before every K characters. Finally, return the new string.
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class Solution { public String licenseKeyFormatting(String S, int K) { S = S.toUpperCase(); S = S.replaceAll("-", ""); int length = S.length(); StringBuffer sb = new StringBuffer(S); int count = 0; for (int i = length - 1; i > 0; i--) { count++; if (count == K) { sb.insert(i, '-'); count = 0; } } return sb.toString(); } } -
Todo -
class Solution(object): def licenseKeyFormatting(self, S, K): """ :type S: str :type K: int :rtype: str """ s = S.split("-") s = "".join(s) n = len(s) start = n % K res = [] if start != 0: res.append(s[:start].upper()) for k in range(0, (len(s) - start) / K): res.append(s[start:start + K].upper()) start += K return "-".join(res)