482. License Key Formatting¶

#
# @lc app=leetcode id=482 lang=python3
#
# [482] License Key Formatting
#
# https://leetcode.com/problems/license-key-formatting/description/
#
# algorithms
# Easy (43.31%)
# Likes:    1032
# Dislikes: 1353
# Total Accepted:    260.6K
# Total Submissions: 600.6K
# Testcase Example:  '"5F3Z-2e-9-w"\n4'
#
# You are given a license key represented as a string s that consists of only
# alphanumeric characters and dashes. The string is separated into n + 1 groups
# by n dashes. You are also given an integer k.
#
# We want to reformat the string s 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 you should convert all lowercase letters to
# uppercase.
#
# Return the reformatted license key.
#
#
# 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.
#
#
#
# Constraints:
#
#
# 1 <= s.length <= 10^5
# s consists of English letters, digits, and dashes '-'.
# 1 <= k <= 10^4
#
#
#

# @lc code=start
class Solution:
    def licenseKeyFormatting(self, s: str, k: int) -> str:
        r = ""

        n = 0
        for i in range(-1, -len(s)-1, -1):
            c = s[i].upper()

            if n % k == 0 and \
                    n != 0 and r[0] != '-':
                r = '-' + r

            if c != '-':
                r = c + r
                n += 1

        return r.strip('-')

# @lc code=end