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Commit 27aed52

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Sync LeetCode submission Runtime - 161 ms (71.55%), Memory - 36 MB (57.76%)
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‎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>
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<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>
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<ul>
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<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>
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</ul>
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<p>&nbsp;</p>
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<p><strong class="example">Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> words = [&quot;area&quot;,&quot;lead&quot;,&quot;wall&quot;,&quot;lady&quot;,&quot;ball&quot;]
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<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;]]
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<strong>Explanation:</strong>
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The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).
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</pre>
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<p><strong class="example">Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> words = [&quot;abat&quot;,&quot;baba&quot;,&quot;atan&quot;,&quot;atal&quot;]
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<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;]]
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<strong>Explanation:</strong>
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The output consists of two word squares. The order of output does not matter (just the order of words in each word square matters).
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= words.length &lt;= 1000</code></li>
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<li><code>1 &lt;= words[i].length &lt;= 4</code></li>
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<li>All <code>words[i]</code> have the same length.</li>
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<li><code>words[i]</code> consists of only lowercase English letters.</li>
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<li>All <code>words[i]</code> are <strong>unique</strong>.</li>
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</ul>

‎0425-word-squares/solution.py‎

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

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