Solution of Next Time and License Key

Author: Garvit244

400-500Q/482.py

@@ -0,0 +1,44 @@
+'''
+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.
+'''
+
+class Solution(object):
+    def licenseKeyFormatting(self, S, K):
+        """
+        :type S: str
+        :type K: int
+        :rtype: str
+        """
+        S = S.replace('-','').upper()
+        result = ""
+        
+        if len(S)%K == 0:
+            for index in range(0, len(S), K):
+                result += S[index:index+K] + "-"
+        else:
+            result = S[:len(S)%K] + "-"
+            for index in range(len(S)%K, len(S), K):
+                result += S[index:index+K] + "-"
+        return result[:-1]

600-700q/681.py

@@ -0,0 +1,33 @@
+'''
+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.
+
+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.
+
+Example 1:
+
+Input: "19:34"
+Output: "19:39"
+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.
+Example 2:
+
+Input: "23:59"
+Output: "22:22"
+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.
+'''
+
+class Solution(object):
+    def nextClosestTime(self, time):
+    	current_time = 60*int(time[:2]) + int(time[3:])
+    	allowed = {int(x) for x in time if x != ':'}
+    	result = 24*60
+    	ans = current_time
+    	for h1, h2, m1, m2 in itertools.product(allowed, repeat=4):
+    		hours, minutes = 10*h1+h2, 10*m1+m2
+    		if hours < 24 and minutes < 60:
+    			elapsed = 60*hours + minutes
+    			diff = (current_time - elapsed)%(24*60)
+    			if 0 < diff < result:
+    				result = diff
+    				ans = elapsed
+
+    	return "{:02d}:{:02d}".format(divmod(ans, 60))

README.md

@@ -85,13 +85,15 @@ Python solution of problems from [LeetCode](https://leetcode.com/).
 ##### [Problems 600-700](./600-700q/)
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|681|[Next Closest Time ](https://leetcode.com/problems/next-closest-time)|[Python](./600-700q/681.py)|Medium|
 |674|[Longest Continuous Increasing Subsequence](https://leetcode.com/problems/longest-continuous-increasing-subsequence)|[Python](./600-700/674.py)|Easy|
 |673|[Number of Longest Increasing Subsequence](https://leetcode.com/problems/number-of-longest-increasing-subsequence)|[Python](./600-700q/673.py)|Medium|
 
 
 ##### [Problems 400-500](./400-500Q/)
 | # | Title | Solution | Difficulty |
 |---| ----- | -------- | ---------- |
+|482|[License Key Formatting](https://leetcode.com/problems/license-key-formatting)|[Python](./400-500q/482.py)|Easy|
 |454|[4Sum II](https://leetcode.com/problems/4sum-ii/)|[Python](./400-500Q/454.py)|Medium|
 |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|
 |442|[Find All Duplicates in an Array](https://leetcode.com/problems/find-all-duplicates-in-an-array)|[Python](./400-500q/442.py)|Easy|