License Key Formatting
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.
- 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.
Credits To: leetcode.com
Leave me comments, if you have better ways to solve.
## Blog link: https://code.dennyzhang.com/license-key-formatting ## Basic Ideas: Get the length of non-dash string ## Then we know how many dash we need ## Move from right to left with two pointers ## Complexity: Time O(n), Space O(n) (If list instead of string, we can solve O(1) space) class Solution(object): def licenseKeyFormatting(self, S, K): """ :type S: str :type K: int :rtype: str """ length = len(S) count_str = length - S.count('-') count_group = count_str/K if count_str % K != 0: count_group += 1 l = [None] * (count_str + count_group - 1) # get result from the right to left index, count = len(l)-1, K for i in xrange(length-1, -1, -1): if index == -1: break if count == 0: l[index] = '-' index, count = index-1, K ch = S[i] if ch != '-': l[index] = ch.upper() index, count = index-1, count-1 return ''.join(l) # s = Solution() # s.licenseKeyFormatting("--a-a-a-a--", 2)