<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:
        # i: the current index for pattern
        # pick a starting point
        # for start in range(1, 10)

        def backtrack(i, path, ans, taken):
            if i >= len(pattern):
                if ''.join(path) < ans[0]:
                    ans[0] = ''.join(path)
                return

            if pattern[i] == 'I':
                for j in range(1, 10):
                    if not taken[j]:
                        if len(path) == 0 or int(path[len(path)-1]) < j: 
                            taken[j] = True
                            path.append(str(j))
                            backtrack(i+1, path, ans, taken)
                            path.pop()
                            taken[j] = False
            elif pattern[i] == 'D':
                for j in range(1,10):
                    if not taken[j]:
                        if len(path) == 0 or int(path[len(path)-1]) > j:
                            taken[j] = True
                            path.append(str(j))
                            backtrack(i+1, path, ans, taken)
                            path.pop()
                            taken[j] = False
            
        taken = [False] * 10
        ans = [''.join(['9']*9)]
        path = []
        for i in range(1, 10):
            taken[i] = True
            path.append(str(i))
            backtrack(0, path, ans, taken)
            path.pop()
            taken[i] = False
        return ans[0]

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