Sync LeetCode submission Runtime - 507 ms (34.66%), Memory - 21 MB (45.02%)

Author: jimit105

0952-word-subsets/README.md

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+<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>&quot;wrr&quot;</code> is a subset of <code>&quot;warrior&quot;</code> but is not a subset of <code>&quot;world&quot;</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 class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> words1 = [&quot;amazon&quot;,&quot;apple&quot;,&quot;facebook&quot;,&quot;google&quot;,&quot;leetcode&quot;], words2 = [&quot;e&quot;,&quot;o&quot;]
+<strong>Output:</strong> [&quot;facebook&quot;,&quot;google&quot;,&quot;leetcode&quot;]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> words1 = [&quot;amazon&quot;,&quot;apple&quot;,&quot;facebook&quot;,&quot;google&quot;,&quot;leetcode&quot;], words2 = [&quot;l&quot;,&quot;e&quot;]
+<strong>Output:</strong> [&quot;apple&quot;,&quot;google&quot;,&quot;leetcode&quot;]
+</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>

0952-word-subsets/solution.py

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+# Approach 1: Reduce to Single Word in B
+
+# A = len(words1) ,B = len(words2), and W is max word length
+# Time: O(A * W + B * W)
+# Space: O(A + B) -> O(A)
+
+class Solution:
+    def wordSubsets(self, words1: List[str], words2: List[str]) -> List[str]:
+        def count(word):
+            ans = [0] * 26
+            for letter in word:
+                ans[ord(letter) - ord('a')] += 1
+            return ans
+
+        bmax = [0] * 26
+        for b in words2:
+            for i, c in enumerate(count(b)):
+                bmax[i] = max(bmax[i], c)
+
+        ans = []
+        for a in words1:
+            if all(x >= y for x, y in zip(count(a), bmax)):
+                ans.append(a)
+
+        return ans
+