## 748. Shortest Completing Word

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:

1. `licensePlate` will be a string with length in range `[1, 7]`.
2. `licensePlate` will contain digits, spaces, or letters (uppercase or lowercase).
3. `words` will have a length in the range `[10, 1000]`.
4. Every `words[i]` will consist of lowercase letters, and have length in range `[1, 15]`.

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#### Approach #1: Compare Counts [Accepted]

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Intuition and Algorithm

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A natural question is, how to tell whether a `word` like `"steps"` completes a `licensePlate` like `"12s pst"`?

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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`.

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From the words that complete `licensePlate`, we should keep the one with the shortest length (with ties broken by whether it occurs first.)

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Complexity Analysis

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Time Complexity: where is the length of `words`, and assuming the lengths of `licensePlate` and `words[i]` are bounded by .

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