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Commit 9a2d5db

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Solution of Next Time and License Key
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‎400-500Q/482.py‎

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'''
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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.
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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.
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Given a non-empty string S and a number K, format the string according to the rules described above.
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Example 1:
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Input: S = "5F3Z-2e-9-w", K = 4
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Output: "5F3Z-2E9W"
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Explanation: The string S has been split into two parts, each part has 4 characters.
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Note that the two extra dashes are not needed and can be removed.
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Example 2:
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Input: S = "2-5g-3-J", K = 2
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Output: "2-5G-3J"
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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.
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Note:
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The length of string S will not exceed 12,000, and K is a positive integer.
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String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
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String S is non-empty.
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'''
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class Solution(object):
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def licenseKeyFormatting(self, S, K):
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"""
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:type S: str
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:type K: int
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:rtype: str
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"""
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S = S.replace('-','').upper()
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result = ""
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if len(S)%K == 0:
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for index in range(0, len(S), K):
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result += S[index:index+K] + "-"
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else:
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result = S[:len(S)%K] + "-"
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for index in range(len(S)%K, len(S), K):
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result += S[index:index+K] + "-"
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return result[:-1]

‎600-700q/681.py‎

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'''
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Given a time represented in the format "HH:MM", form the next closest time by reusing the current digits. There is no limit on how many times a digit can be reused.
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You may assume the given input string is always valid. For example, "01:34", "12:09" are all valid. "1:34", "12:9" are all invalid.
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Example 1:
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Input: "19:34"
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Output: "19:39"
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Explanation: The next closest time choosing from digits 1, 9, 3, 4, is 19:39, which occurs 5 minutes later. It is not 19:33, because this occurs 23 hours and 59 minutes later.
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Example 2:
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Input: "23:59"
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Output: "22:22"
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Explanation: The next closest time choosing from digits 2, 3, 5, 9, is 22:22. It may be assumed that the returned time is next day's time since it is smaller than the input time numerically.
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'''
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class Solution(object):
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def nextClosestTime(self, time):
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current_time = 60*int(time[:2]) + int(time[3:])
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allowed = {int(x) for x in time if x != ':'}
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result = 24*60
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ans = current_time
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for h1, h2, m1, m2 in itertools.product(allowed, repeat=4):
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hours, minutes = 10*h1+h2, 10*m1+m2
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if hours < 24 and minutes < 60:
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elapsed = 60*hours + minutes
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diff = (current_time - elapsed)%(24*60)
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if 0 < diff < result:
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result = diff
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ans = elapsed
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return "{:02d}:{:02d}".format(divmod(ans, 60))

‎README.md‎

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##### [Problems 600-700](./600-700q/)
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| # | Title | Solution | Difficulty |
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|---| ----- | -------- | ---------- |
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|681|[Next Closest Time ](https://leetcode.com/problems/next-closest-time)|[Python](./600-700q/681.py)|Medium|
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|674|[Longest Continuous Increasing Subsequence](https://leetcode.com/problems/longest-continuous-increasing-subsequence)|[Python](./600-700/674.py)|Easy|
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|673|[Number of Longest Increasing Subsequence](https://leetcode.com/problems/number-of-longest-increasing-subsequence)|[Python](./600-700q/673.py)|Medium|
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##### [Problems 400-500](./400-500Q/)
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| # | Title | Solution | Difficulty |
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|---| ----- | -------- | ---------- |
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|482|[License Key Formatting](https://leetcode.com/problems/license-key-formatting)|[Python](./400-500q/482.py)|Easy|
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|454|[4Sum II](https://leetcode.com/problems/4sum-ii/)|[Python](./400-500Q/454.py)|Medium|
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|448|[Find All Numbers Disappeared in an Array](https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array)|[Python](./400-500q/448.py)|Easy|
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|442|[Find All Duplicates in an Array](https://leetcode.com/problems/find-all-duplicates-in-an-array)|[Python](./400-500q/442.py)|Easy|

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