Sync LeetCode submission Runtime - 298 ms (66.67%), Memory - 18.1 MB (93.97%)

Author: jimit105

1302-delete-characters-to-make-fancy-string/README.md

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+<p>A <strong>fancy string</strong> is a string where no <strong>three</strong> <strong>consecutive</strong> characters are equal.</p>
+
+<p>Given a string <code>s</code>, delete the <strong>minimum</strong> possible number of characters from <code>s</code> to make it <strong>fancy</strong>.</p>
+
+<p>Return <em>the final string after the deletion</em>. It can be shown that the answer will always be <strong>unique</strong>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;le<u>e</u>etcode&quot;
+<strong>Output:</strong> &quot;leetcode&quot;
+<strong>Explanation:</strong>
+Remove an &#39;e&#39; from the first group of &#39;e&#39;s to create &quot;leetcode&quot;.
+No three consecutive characters are equal, so return &quot;leetcode&quot;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;<u>a</u>aab<u>aa</u>aa&quot;
+<strong>Output:</strong> &quot;aabaa&quot;
+<strong>Explanation:</strong>
+Remove an &#39;a&#39; from the first group of &#39;a&#39;s to create &quot;aabaaaa&quot;.
+Remove two &#39;a&#39;s from the second group of &#39;a&#39;s to create &quot;aabaa&quot;.
+No three consecutive characters are equal, so return &quot;aabaa&quot;.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;aab&quot;
+<strong>Output:</strong> &quot;aab&quot;
+<strong>Explanation:</strong> No three consecutive characters are equal, so return &quot;aab&quot;.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>s</code> consists only of lowercase English letters.</li>
+</ul>

1302-delete-characters-to-make-fancy-string/solution.py

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+# Approach 1: Insert characters in a new string
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def makeFancyString(self, s: str) -> str:
+        prev = s[0]
+        freq = 1
+        ans = s[0]
+
+        for i in range(1, len(s)):
+            if s[i] == prev:
+                freq += 1
+            else:
+                prev = s[i]
+                freq = 1
+
+            if freq < 3:
+                ans += s[i]
+
+        return ans