Create shortest-completing-word.py

Author: kamyu104

Python/shortest-completing-word.py

@@ -0,0 +1,53 @@
+# Time:  O(n)
+# Space: O(1)
+
+# Find the minimum length word from a given dictionary words,
+# which has all the letters from the string licensePlate.
+# Such a word is said to complete the given string licensePlate
+#
+# Here, for letters we ignore case.
+# For example, "P" on the licensePlate still matches "p" on the word.
+#
+# It is guaranteed an answer exists.
+# If there are multiple answers, return the one that occurs first in the array.
+#
+# The license plate might have the same letter occurring multiple times.
+# For example, given a licensePlate of "PP",
+# the word "pair" does not complete the licensePlate, but the word "supper" does.
+#
+# Example 1:
+# Input: licensePlate = "1s3 PSt", words = ["step", "steps", "stripe", "stepple"]
+# Output: "steps"
+# Explanation: The smallest length word that contains the letters "S", "P", "S", and "T".
+# Note that the answer is not "step", because the letter "s" must occur in the word twice.
+# Also note that we ignored case for the purposes of comparing whether a letter exists in the word.
+# Example 2:
+# Input: licensePlate = "1s3 456", words = ["looks", "pest", "stew", "show"]
+# Output: "pest"
+# Explanation: There are 3 smallest length words that contains the letters "s".
+# We return the one that occurred first.
+# Note:
+# - licensePlate will be a string with length in range [1, 7].
+# - licensePlate will contain digits, spaces, or letters (uppercase or lowercase).
+# - words will have a length in the range [10, 1000].
+# - Every words[i] will consist of lowercase letters, and have length in range [1, 15].
+
+class Solution(object):
+    def shortestCompletingWord(self, licensePlate, words):
+        """
+        :type licensePlate: str
+        :type words: List[str]
+        :rtype: str
+        """
+        def contains(counter1, w2):
+            c2 = collections.Counter(w2.lower())
+            c2.subtract(counter1)
+            return all(map(lambda x: x >= 0, c2.values()))
+
+        result = None
+        counter = collections.Counter(c.lower() for c in licensePlate if c.isalpha())
+        for word in words:
+            if (result is None or (len(word) < len(result))) and \
+               contains(counter, word):
+                result = word
+        return result