Text Justification/ Minimum Window Substring

Author: gavinfish

068 Text Justification.py

@@ -0,0 +1,58 @@
+'''
+Given an array of words and a length L, format the text such that each line has exactly L characters and is fully (left and right) justified.
+
+You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad extra spaces ' ' when necessary so that each line has exactly L characters.
+
+Extra spaces between words should be distributed as evenly as possible. If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right.
+
+For the last line of text, it should be left justified and no extra space is inserted between words.
+
+For example,
+words: ["This", "is", "an", "example", "of", "text", "justification."]
+L: 16.
+
+Return the formatted lines as:
+[
+   "This    is    an",
+   "example  of text",
+   "justification.  "
+]
+Note: Each word is guaranteed not to exceed L in length.
+
+A line other than the last line might contain only one word. What should you do in this case?
+In this case, that line should be left-justified.
+'''
+
+class Solution(object):
+    def fullJustify(self, words, maxWidth):
+        """
+        :type words: List[str]
+        :type maxWidth: int
+        :rtype: List[str]
+        """
+        start = end = 0
+        result, curr_words_length = [], 0
+        for i, word in enumerate(words):
+            if len(word) + curr_words_length + end - start  > maxWidth:
+                if end - start == 1:
+                    result.append(words[start] + ' ' * (maxWidth - curr_words_length))
+                else:
+                    total_space = maxWidth - curr_words_length
+                    space, extra = divmod(total_space, end - start - 1)
+                    for j in range(extra):
+                        words[start + j] += ' '
+                    result.append((' ' * space).join(words[start:end]))
+                curr_words_length = 0
+                start = end = i
+            end += 1
+            curr_words_length += len(word)
+        result.append(' '.join(words[start:end]) + ' ' * (maxWidth - curr_words_length - (end - start - 1)))
+        return result
+
+
+if __name__ == "__main__":
+    assert Solution().fullJustify(["This", "is", "an", "example", "of", "text", "justification."], 16) == [
+        "This    is    an",
+        "example  of text",
+        "justification.  "
+    ]

076 Minimum Window Substring.py

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+'''
+Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
+
+For example,
+S = "ADOBECODEBANC"
+T = "ABC"
+Minimum window is "BANC".
+
+Note:
+If there is no such window in S that covers all characters in T, return the empty string "".
+
+If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
+'''
+
+from collections import defaultdict
+
+
+class Solution(object):
+    def minWindow(self, s, t):
+        """
+        :type s: str
+        :type t: str
+        :rtype: str
+        """
+        MAX_INT = 2147483647
+        start = end = 0
+        char_need = defaultdict(int)    # the count of char needed by current window, negative means current window has it but not needs it
+        count_need = len(t)             # count of chars not in current window but in t
+        min_length = MAX_INT
+        min_start = 0
+        for i in t:
+            char_need[i] += 1           # current window needs all char in t
+        while end < len(s):
+            if char_need[s[end]] > 0:
+                count_need -= 1
+            char_need[s[end]] -= 1      # current window contains s[end] now, so does not need it any more
+            end += 1
+            while count_need == 0:
+                if min_length > end - start:
+                    min_length = end - start
+                    min_start = start
+                char_need[s[start]] += 1    # current window does not contain s[start] any more
+                if char_need[s[start]] > 0: # when some count in char_need is positive, it means there is char in t but not current window
+                    count_need += 1
+                start += 1
+        return "" if min_length == MAX_INT else s[min_start:min_start + min_length]
+
+
+if __name__ == "__main__":
+    assert Solution().minWindow("ADOBECODEBANC", "ABC") == "BANC"