<p>You are given a <strong>0-indexed</strong> string <code>pattern</code> of length <code>n</code> consisting of the characters <code>&#39;I&#39;</code> meaning <strong>increasing</strong> and <code>&#39;D&#39;</code> meaning <strong>decreasing</strong>.</p>

<p>A <strong>0-indexed</strong> string <code>num</code> of length <code>n + 1</code> is created using the following conditions:</p>

<ul>
	<li><code>num</code> consists of the digits <code>&#39;1&#39;</code> to <code>&#39;9&#39;</code>, where each digit is used <strong>at most</strong> once.</li>
	<li>If <code>pattern[i] == &#39;I&#39;</code>, then <code>num[i] &lt; num[i + 1]</code>.</li>
	<li>If <code>pattern[i] == &#39;D&#39;</code>, then <code>num[i] &gt; num[i + 1]</code>.</li>
</ul>

<p>Return <em>the lexicographically <strong>smallest</strong> possible string </em><code>num</code><em> that meets the conditions.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> pattern = &quot;IIIDIDDD&quot;
<strong>Output:</strong> &quot;123549876&quot;
<strong>Explanation:
</strong>At indices 0, 1, 2, and 4 we must have that num[i] &lt; num[i+1].
At indices 3, 5, 6, and 7 we must have that num[i] &gt; num[i+1].
Some possible values of num are &quot;245639871&quot;, &quot;135749862&quot;, and &quot;123849765&quot;.
It can be proven that &quot;123549876&quot; is the smallest possible num that meets the conditions.
Note that &quot;123414321&quot; is not possible because the digit &#39;1&#39; is used more than once.</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> pattern = &quot;DDD&quot;
<strong>Output:</strong> &quot;4321&quot;
<strong>Explanation:</strong>
Some possible values of num are &quot;9876&quot;, &quot;7321&quot;, and &quot;8742&quot;.
It can be proven that &quot;4321&quot; is the smallest possible num that meets the conditions.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= pattern.length &lt;= 8</code></li>
	<li><code>pattern</code> consists of only the letters <code>&#39;I&#39;</code> and <code>&#39;D&#39;</code>.</li>
</ul>

class Solution:
    def smallestNumber(self, pattern: str) -> str:
        n = len(pattern)

        @cache
        def dp(i, cur):
            if i == n:
                return int(cur)
            res = 999999999
            d = pattern[i]
            if d == "I":
                for nex in range(int(cur[-1])+1, 10):
                    if str(nex) not in cur:
                        res = min(res, dp(i+1, cur+str(nex)))
            else:
                for nex in range(1, int(cur[-1])):
                    if str(nex) not in cur:
                        res = min(res, dp(i+1, cur+str(nex)))
            return res


        return str(min(dp(0, str(i)) for i in range(1,10)))
473