Leetcode: License Key Formatting

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.

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.

Github: code.dennyzhang.com

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)
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