Construct Smallest Number From DI String
<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>
Stack; Greedy; String; Backtracking
class Solution:
    def smallestNumber(self, pattern: str) -> str:
        def dfs(u):
            nonlocal ans
            if ans:
                return
            if u == len(pattern) + 1:
                ans = ''.join(t)
                return
            for i in range(1, 10):
                if not vis[i]:
                    if u and pattern[u - 1] == 'I' and int(t[-1]) >= i:
                        continue
                    if u and pattern[u - 1] == 'D' and int(t[-1]) <= i:
                        continue
                    vis[i] = True
                    t.append(str(i))
                    dfs(u + 1)
                    vis[i] = False
                    t.pop()

        vis = [False] * 10
        t = []
        ans = None
        dfs(0)
        return ans
