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Author: jimit105

0807-custom-sort-string/README.md

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+<p>You are given two strings <code>order</code> and <code>s</code>. All the characters of <code>order</code> are <strong>unique</strong> and were sorted in some custom order previously.</p>
+
+<p>Permute the characters of <code>s</code> so that they match the order that <code>order</code> was sorted. More specifically, if a character <code>x</code> occurs before a character <code>y</code> in <code>order</code>, then <code>x</code> should occur before <code>y</code> in the permuted string.</p>
+
+<p>Return <em>any permutation of </em><code>s</code><em> that satisfies this property</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> order = &quot;cba&quot;, s = &quot;abcd&quot; </span></p>
+
+<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> &quot;cbad&quot; </span></p>
+
+<p><strong>Explanation: </strong> <code>&quot;a&quot;</code>, <code>&quot;b&quot;</code>, <code>&quot;c&quot;</code> appear in order, so the order of <code>&quot;a&quot;</code>, <code>&quot;b&quot;</code>, <code>&quot;c&quot;</code> should be <code>&quot;c&quot;</code>, <code>&quot;b&quot;</code>, and <code>&quot;a&quot;</code>.</p>
+
+<p>Since <code>&quot;d&quot;</code> does not appear in <code>order</code>, it can be at any position in the returned string. <code>&quot;dcba&quot;</code>, <code>&quot;cdba&quot;</code>, <code>&quot;cbda&quot;</code> are also valid outputs.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
+<p><strong>Input: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> order = &quot;bcafg&quot;, s = &quot;abcd&quot; </span></p>
+
+<p><strong>Output: </strong> <span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;"> &quot;bcad&quot; </span></p>
+
+<p><strong>Explanation: </strong> The characters <code>&quot;b&quot;</code>, <code>&quot;c&quot;</code>, and <code>&quot;a&quot;</code> from <code>order</code> dictate the order for the characters in <code>s</code>. The character <code>&quot;d&quot;</code> in <code>s</code> does not appear in <code>order</code>, so its position is flexible.</p>
+
+<p>Following the order of appearance in <code>order</code>, <code>&quot;b&quot;</code>, <code>&quot;c&quot;</code>, and <code>&quot;a&quot;</code> from <code>s</code> should be arranged as <code>&quot;b&quot;</code>, <code>&quot;c&quot;</code>, <code>&quot;a&quot;</code>. <code>&quot;d&quot;</code> can be placed at any position since it&#39;s not in order. The output <code>&quot;bcad&quot;</code> correctly follows this rule. Other arrangements like <code>&quot;dbca&quot;</code> or <code>&quot;bcda&quot;</code> would also be valid, as long as <code>&quot;b&quot;</code>, <code>&quot;c&quot;</code>, <code>&quot;a&quot;</code> maintain their order.</p>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= order.length &lt;= 26</code></li>
+	<li><code>1 &lt;= s.length &lt;= 200</code></li>
+	<li><code>order</code> and <code>s</code> consist of lowercase English letters.</li>
+	<li>All the characters of <code>order</code> are <strong>unique</strong>.</li>
+</ul>

0807-custom-sort-string/solution.py

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+# Approach 2: Frequency Table and Counting
+
+# m = length of `order`, n = length of `s`
+# Time: O(m + n)
+# Space: O(n)
+
+from collections import Counter
+
+class Solution:
+    def customSortString(self, order: str, s: str) -> str:
+        char_counts = Counter(s)
+        result = ''
+
+        for char in order:
+            if char in char_counts:
+                result += char * char_counts[char]
+                del char_counts[char]
+
+        result += ''.join(char * count for char, count in char_counts.items())
+
+        return result
+