Sync LeetCode submission Runtime - 0 ms (100.00%), Memory - 19.3 MB (40.87%)

Author: koralkulacoglu

LeetCode/2093-check-if-string-is-a-prefix-of-array/README.md

@@ -0,0 +1,33 @@
+<p>Given a string <code>s</code> and an array of strings <code>words</code>, determine whether <code>s</code> is a <strong>prefix string</strong> of <code>words</code>.</p>
+
+<p>A string <code>s</code> is a <strong>prefix string</strong> of <code>words</code> if <code>s</code> can be made by concatenating the first <code>k</code> strings in <code>words</code> for some <strong>positive</strong> <code>k</code> no larger than <code>words.length</code>.</p>
+
+<p>Return <code>true</code><em> if </em><code>s</code><em> is a <strong>prefix string</strong> of </em><code>words</code><em>, or </em><code>false</code><em> otherwise</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;iloveleetcode&quot;, words = [&quot;i&quot;,&quot;love&quot;,&quot;leetcode&quot;,&quot;apples&quot;]
+<strong>Output:</strong> true
+<strong>Explanation:</strong>
+s can be made by concatenating &quot;i&quot;, &quot;love&quot;, and &quot;leetcode&quot; together.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;iloveleetcode&quot;, words = [&quot;apples&quot;,&quot;i&quot;,&quot;love&quot;,&quot;leetcode&quot;]
+<strong>Output:</strong> false
+<strong>Explanation:</strong>
+It is impossible to make s using a prefix of arr.</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;= 20</code></li>
+	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
+	<li><code>words[i]</code> and <code>s</code> consist of only lowercase English letters.</li>
+</ul>

LeetCode/2093-check-if-string-is-a-prefix-of-array/solution.py

@@ -0,0 +1,10 @@
+class Solution:
+    def isPrefixString(self, s: str, words: List[str]) -> bool:
+        for word in words:
+            if len(s) == 0:
+                return True
+            if s.startswith(word):
+                s = s[len(word):]
+            else:
+                return False
+        return len(s) == 0