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]
.words[i]
will consist of lowercase letters, and have length in range [1, 15]
.Intuition and Algorithm
\nA natural question is, how to tell whether a word
like "steps"
completes a licensePlate
like "12s pst"
?
We count the number of letters in both word
and licensePlate
, converting to lowercase and ignoring non-letter characters. If the count of each letter is greater or equal in the word, then that word completes the licensePlate
.
From the words that complete licensePlate
, we should keep the one with the shortest length (with ties broken by whether it occurs first.)
Complexity Analysis
\nTime Complexity: where is the length of words
, and assuming the lengths of licensePlate
and words[i]
are bounded by .
Space Complexity: in additional space.
\nAnalysis written by: @awice.
\n