# 916-word-subsets


Try it on <a href='https://leetcode.com/problems/916-word-subsets'>leetcode</a>

## Description
<div class="description">
<div><p>You are given two string arrays <code>words1</code> and <code>words2</code>.</p>

<p>A string <code>b</code> is a <strong>subset</strong> of string <code>a</code> if every letter in <code>b</code> occurs in <code>a</code> including multiplicity.</p>

<ul>
	<li>For example, <code>"wrr"</code> is a subset of <code>"warrior"</code> but is not a subset of <code>"world"</code>.</li>
</ul>

<p>A string <code>a</code> from <code>words1</code> is <strong>universal</strong> if for every string <code>b</code> in <code>words2</code>, <code>b</code> is a subset of <code>a</code>.</p>

<p>Return an array of all the <strong>universal</strong> strings in <code>words1</code>. You may return the answer in <strong>any order</strong>.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre><strong>Input:</strong> words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["e","o"]
<strong>Output:</strong> ["facebook","google","leetcode"]
</pre>

<p><strong>Example 2:</strong></p>

<pre><strong>Input:</strong> words1 = ["amazon","apple","facebook","google","leetcode"], words2 = ["l","e"]
<strong>Output:</strong> ["apple","google","leetcode"]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= words1.length, words2.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words1[i].length, words2[i].length &lt;= 10</code></li>
	<li><code>words1[i]</code> and <code>words2[i]</code> consist only of lowercase English letters.</li>
	<li>All the strings of <code>words1</code> are <strong>unique</strong>.</li>
</ul>
</div>
</div>

## Solution(Python)
```Python
class Solution(object):
    def wordSubsets(self, A, B):
        def count(word):
            ans = [0] * 26
            for letter in word:
                ans[ord(letter) - ord('a')] += 1
            return ans

        bmax = [0] * 26
        for b in B:
            for i, c in enumerate(count(b)):
                bmax[i] = max(bmax[i], c)

        ans = []
        for a in A:
            if all(x >= y for x, y in zip(count(a), bmax)):
                ans.append(a)
        return ans
```