<p>An original string, consisting of lowercase English letters, can be encoded by the following steps:</p>

<ul>
	<li>Arbitrarily <strong>split</strong> it into a <strong>sequence</strong> of some number of <strong>non-empty</strong> substrings.</li>
	<li>Arbitrarily choose some elements (possibly none) of the sequence, and <strong>replace</strong> each with <strong>its length</strong> (as a numeric string).</li>
	<li><strong>Concatenate</strong> the sequence as the encoded string.</li>
</ul>

<p>For example, <strong>one way</strong> to encode an original string <code>&quot;abcdefghijklmnop&quot;</code> might be:</p>

<ul>
	<li>Split it as a sequence: <code>[&quot;ab&quot;, &quot;cdefghijklmn&quot;, &quot;o&quot;, &quot;p&quot;]</code>.</li>
	<li>Choose the second and third elements to be replaced by their lengths, respectively. The sequence becomes <code>[&quot;ab&quot;, &quot;12&quot;, &quot;1&quot;, &quot;p&quot;]</code>.</li>
	<li>Concatenate the elements of the sequence to get the encoded string: <code>&quot;ab121p&quot;</code>.</li>
</ul>

<p>Given two encoded strings <code>s1</code> and <code>s2</code>, consisting of lowercase English letters and digits <code>1-9</code> (inclusive), return <code>true</code><em> if there exists an original string that could be encoded as <strong>both</strong> </em><code>s1</code><em> and </em><code>s2</code><em>. Otherwise, return </em><code>false</code>.</p>

<p><strong>Note</strong>: The test cases are generated such that the number of consecutive digits in <code>s1</code> and <code>s2</code> does not exceed <code>3</code>.</p>

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

<pre>
<strong>Input:</strong> s1 = &quot;internationalization&quot;, s2 = &quot;i18n&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> It is possible that &quot;internationalization&quot; was the original string.
- &quot;internationalization&quot; 
  -&gt; Split:       [&quot;internationalization&quot;]
  -&gt; Do not replace any element
  -&gt; Concatenate:  &quot;internationalization&quot;, which is s1.
- &quot;internationalization&quot;
  -&gt; Split:       [&quot;i&quot;, &quot;nternationalizatio&quot;, &quot;n&quot;]
  -&gt; Replace:     [&quot;i&quot;, &quot;18&quot;,                 &quot;n&quot;]
  -&gt; Concatenate:  &quot;i18n&quot;, which is s2
</pre>

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

<pre>
<strong>Input:</strong> s1 = &quot;l123e&quot;, s2 = &quot;44&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> It is possible that &quot;leetcode&quot; was the original string.
- &quot;leetcode&quot; 
  -&gt; Split:      [&quot;l&quot;, &quot;e&quot;, &quot;et&quot;, &quot;cod&quot;, &quot;e&quot;]
  -&gt; Replace:    [&quot;l&quot;, &quot;1&quot;, &quot;2&quot;,  &quot;3&quot;,   &quot;e&quot;]
  -&gt; Concatenate: &quot;l123e&quot;, which is s1.
- &quot;leetcode&quot; 
  -&gt; Split:      [&quot;leet&quot;, &quot;code&quot;]
  -&gt; Replace:    [&quot;4&quot;,    &quot;4&quot;]
  -&gt; Concatenate: &quot;44&quot;, which is s2.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s1 = &quot;a5b&quot;, s2 = &quot;c5b&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong> It is impossible.
- The original string encoded as s1 must start with the letter &#39;a&#39;.
- The original string encoded as s2 must start with the letter &#39;c&#39;.
</pre>

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

<ul>
	<li><code>1 &lt;= s1.length, s2.length &lt;= 40</code></li>
	<li><code>s1</code> and <code>s2</code> consist of digits <code>1-9</code> (inclusive), and lowercase English letters only.</li>
	<li>The number of consecutive digits in <code>s1</code> and <code>s2</code> does not exceed <code>3</code>.</li>
</ul>

from functools import cache

class Solution:
    def possiblyEquals(self, s1: str, s2: str) -> bool:
        repr1 = Solution.parse(s1)
        repr2 = Solution.parse(s2)
        i = j = 0
        
        @cache
        def dfs(i, j, diff):
            # if both strings have been exhausted, we're done
            # if only one string is at max-length, then we're probably iterating through
            # the diffs, so we can wait
            if i == len(repr1) and j == len(repr2):
                return diff == 0
            
            # if only one string has been exhausted, we're not able to match the two
            if i > len(repr1) or j > len(repr2):
                return False
            
            ei = repr1[i] if i < len(repr1) else ''
            ej = repr2[j] if j < len(repr2) else ''

            # if one is wildcards and the other is a string
            # go through all the possibilities for wildcards and recurse
            if type(ei) != str and type(ej) == str:
                for di in ei:
                    if dfs(i + 1, j, diff + di):
                        return True
                return False
            
            if type(ei) == str and type(ej) != str:
                for dj in ej:
                    if dfs(i, j + 1, diff - dj):
                        return True
                return False

            # if both are wildcards
            # you need to go through all the combinations
            if type(ei) != str and type(ej) != str:
                for di in ei:
                    for dj in ej:
                        if dfs(i + 1, j + 1, diff + di - dj):
                            return True
                return False
            
            # if both are strings
            if type(ei) == str and type(ej) == str:
                # if you have no wildcards, you're just checking if both strings are equivalent
                if diff == 0:
                    if ei != ej:
                        return False
                    else:
                        return dfs(i + 1, j + 1, diff)
                else:
                    # use up some wildcards
                    if diff > 0:
                        return dfs(i, j + 1, diff - 1)
                    else:
                        return dfs(i + 1, j, diff + 1)
            
        
        return dfs(0, 0, 0)
        
    @staticmethod
    def parse(string):
        res = []
        numbers = []
        for c in string:
            if c.isdigit():
                numbers.append(c)
            else:
                if numbers:
                    res.append(Solution.allpossiblewildcardlengths(numbers, 0, set()))
                    numbers = []
                res.append(c)
        if numbers:
            res.append(Solution.allpossiblewildcardlengths(numbers, 0, set()))
        return res
    
    @staticmethod
    def allpossiblewildcardlengths(nums, carryover = 0, options = set()):
        if len(nums) == 0:
            return

        # everything
        for x in range(1, len(nums) + 1):
            prefix = ''.join(nums[:x])
            pint = int(prefix)
            if x == len(nums):
                options.add(carryover + pint)
            else:
                Solution.allpossiblewildcardlengths(nums[x:], carryover + pint, options)

        return options
367