<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>

class Solution:
    def customSortString(self, order: str, s: str) -> str:
        group = defaultdict(list)
        length = len(order)
        ct_s = Counter(s)
        rest = dict()
        for key, value in ct_s.items():
            print(key)
            if key in order:
                indx = order.index(key)
                group[indx].append((key, value))
            if key not in order:
                indx = length
                group[indx].append((key, value))

        s_group = [group[key] for key in sorted(group)]
        print(s_group)
        s = ""
        for ele in s_group:
            for c, t in ele:
                s +=c*t
        return s
16600