Sync LeetCode submission Runtime - 4 ms (28.54%), Memory - 17.7 MB (28.23%)

Author: jimit105

1524-string-matching-in-an-array/README.md

@@ -0,0 +1,39 @@
+<p>Given an array of string <code>words</code>, return <em>all strings in </em><code>words</code><em> that is a <strong>substring</strong> of another word</em>. You can return the answer in <strong>any order</strong>.</p>
+
+<p>A <strong>substring</strong> is a contiguous sequence of characters within a string</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;mass&quot;,&quot;as&quot;,&quot;hero&quot;,&quot;superhero&quot;]
+<strong>Output:</strong> [&quot;as&quot;,&quot;hero&quot;]
+<strong>Explanation:</strong> &quot;as&quot; is substring of &quot;mass&quot; and &quot;hero&quot; is substring of &quot;superhero&quot;.
+[&quot;hero&quot;,&quot;as&quot;] is also a valid answer.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;leetcode&quot;,&quot;et&quot;,&quot;code&quot;]
+<strong>Output:</strong> [&quot;et&quot;,&quot;code&quot;]
+<strong>Explanation:</strong> &quot;et&quot;, &quot;code&quot; are substring of &quot;leetcode&quot;.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;blue&quot;,&quot;green&quot;,&quot;bu&quot;]
+<strong>Output:</strong> []
+<strong>Explanation:</strong> No string of words is substring of another string.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= words.length &lt;= 100</code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 30</code></li>
+	<li><code>words[i]</code> contains only lowercase English letters.</li>
+	<li>All the strings of <code>words</code> are <strong>unique</strong>.</li>
+</ul>

1524-string-matching-in-an-array/solution.py

@@ -0,0 +1,20 @@
+# Approach 1: Brute Force
+
+# n = size of `words` array, m = length of longest string in `words`
+# Time: O(m * n^2)
+# Space: O(m * n)
+
+class Solution:
+    def stringMatching(self, words: List[str]) -> List[str]:
+        matching_words = []
+
+        for curr_word_idx in range(len(words)):
+            for other_word_idx in range(len(words)):
+                if curr_word_idx == other_word_idx:
+                    continue
+                if words[curr_word_idx] in words[other_word_idx]:
+                    matching_words.append(words[curr_word_idx])
+                    break
+
+        return matching_words
+