Sync LeetCode submission Runtime - 161 ms (71.55%), Memory - 36 MB (57.76%)

Author: jimit105

0425-word-squares/README.md

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+<p>Given an array of <strong>unique</strong> strings <code>words</code>, return <em>all the </em><strong><a href="https://en.wikipedia.org/wiki/Word_square" target="_blank">word squares</a></strong><em> you can build from </em><code>words</code>. The same word from <code>words</code> can be used <strong>multiple times</strong>. You can return the answer in <strong>any order</strong>.</p>
+
+<p>A sequence of strings forms a valid <strong>word square</strong> if the <code>k<sup>th</sup></code> row and column read the same string, where <code>0 &lt;= k &lt; max(numRows, numColumns)</code>.</p>
+
+<ul>
+	<li>For example, the word sequence <code>[&quot;ball&quot;,&quot;area&quot;,&quot;lead&quot;,&quot;lady&quot;]</code> forms a word square because each word reads the same both horizontally and vertically.</li>
+</ul>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;area&quot;,&quot;lead&quot;,&quot;wall&quot;,&quot;lady&quot;,&quot;ball&quot;]
+<strong>Output:</strong> [[&quot;ball&quot;,&quot;area&quot;,&quot;lead&quot;,&quot;lady&quot;],[&quot;wall&quot;,&quot;area&quot;,&quot;lead&quot;,&quot;lady&quot;]]
+<strong>Explanation:</strong>
+The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;abat&quot;,&quot;baba&quot;,&quot;atan&quot;,&quot;atal&quot;]
+<strong>Output:</strong> [[&quot;baba&quot;,&quot;abat&quot;,&quot;baba&quot;,&quot;atal&quot;],[&quot;baba&quot;,&quot;abat&quot;,&quot;baba&quot;,&quot;atan&quot;]]
+<strong>Explanation:</strong>
+The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= words.length &lt;= 1000</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 4</code></li>
+	<li>All <code>words[i]</code> have the same length.</li>
+	<li><code>words[i]</code> consists of only lowercase English letters.</li>
+	<li>All <code>words[i]</code> are <strong>unique</strong>.</li>
+</ul>

0425-word-squares/solution.py

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+class Solution:
+
+    def wordSquares(self, words: List[str]) -> List[List[str]]:
+
+        self.words = words
+        self.N = len(words[0])
+        self.buildPrefixHashTable(self.words)
+
+        results = []
+        word_squares = []
+        for word in words:
+            word_squares = [word]
+            self.backtracking(1, word_squares, results)
+        return results
+
+    def backtracking(self, step, word_squares, results):
+        if step == self.N:
+            results.append(word_squares[:])
+            return
+
+        prefix = ''.join([word[step] for word in word_squares])
+        for candidate in self.getWordsWithPrefix(prefix):
+            word_squares.append(candidate)
+            self.backtracking(step+1, word_squares, results)
+            word_squares.pop()
+
+    def buildPrefixHashTable(self, words):
+        self.prefixHashTable = {}
+        for word in words:
+            for prefix in (word[:i] for i in range(1, len(word))):
+                self.prefixHashTable.setdefault(prefix, set()).add(word)
+
+    def getWordsWithPrefix(self, prefix):
+        if prefix in self.prefixHashTable:
+            return self.prefixHashTable[prefix]
+        else:
+            return set([])