Sync LeetCode submission Runtime - 215 ms (39.83%), Memory - 16.8 MB (66.10%)

Author: jimit105

2755-extra-characters-in-a-string/README.md

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+<p>You are given a <strong>0-indexed</strong> string <code>s</code> and a dictionary of words <code>dictionary</code>. You have to break <code>s</code> into one or more <strong>non-overlapping</strong> substrings such that each substring is present in <code>dictionary</code>. There may be some <strong>extra characters</strong> in <code>s</code> which are not present in any of the substrings.</p>
+
+<p>Return <em>the <strong>minimum</strong> number of extra characters left over if you break up </em><code>s</code><em> optimally.</em></p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;leetscode&quot;, dictionary = [&quot;leet&quot;,&quot;code&quot;,&quot;leetcode&quot;]
+<strong>Output:</strong> 1
+<strong>Explanation:</strong> We can break s in two substrings: &quot;leet&quot; from index 0 to 3 and &quot;code&quot; from index 5 to 8. There is only 1 unused character (at index 4), so we return 1.
+
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;sayhelloworld&quot;, dictionary = [&quot;hello&quot;,&quot;world&quot;]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> We can break s in two substrings: &quot;hello&quot; from index 3 to 7 and &quot;world&quot; from index 8 to 12. The characters at indices 0, 1, 2 are not used in any substring and thus are considered as extra characters. Hence, we return 3.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 50</code></li>
+	<li><code>1 &lt;= dictionary.length &lt;= 50</code></li>
+	<li><code>1 &lt;= dictionary[i].length &lt;= 50</code></li>
+	<li><code>dictionary[i]</code>&nbsp;and <code>s</code> consists of only lowercase English letters</li>
+	<li><code>dictionary</code> contains distinct words</li>
+</ul>

2755-extra-characters-in-a-string/solution.py

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+# Approach 1: Top Down Dynamic Programming with Substring Method
+
+# Time: O(N^3)
+
+class Solution:
+    def minExtraChar(self, s: str, dictionary: List[str]) -> int:
+        n, dict_set = len(s), set(dictionary)
+
+        @cache
+        def dp(start):
+            if start == n:
+                return 0
+
+            ans = dp(start + 1) + 1
+            for end in range(start, n):
+                curr = s[start : end + 1]
+                if curr in dict_set:
+                    ans = min(ans, dp(end + 1))
+
+            return ans
+
+        return dp(0)
+