<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>

class Solution:
    def possiblyEquals(self, s1: str, s2: str) -> bool:
        m, n = len(s1), len(s2)
        
        # Tracking state index i and j of s1 and s2, `diff` means the size diff between 2 potential original strings during DFS. For s1, decrease diff, for s2 increase diff. Balance of zero means success in the end.
        @cache
        def DFS(i, j, diff):
            if i == m and j == n:
                return diff == 0
            if i < m and s1[i].isdigit():
                num, i_new = self.get_number(s1, i)
                for count in self.get_all_combination(num):
                    if DFS(i_new, j, diff - count):
                        return True
            elif j < n and s2[j].isdigit():
                num, j_new = self.get_number(s2, j)
                for count in self.get_all_combination(num):
                    if DFS(i, j_new, diff + count):
                        return True
            elif diff == 0: 
                if i < m and j < n and s1[i] == s2[j]: # char needs to match
                    return DFS(i + 1, j + 1, diff)
            elif diff < 0 and j < n: # diff < 0 means s1 is moving ahead
                return DFS(i, j + 1, diff + 1)
            elif diff > 0 and i < m:
                return DFS(i + 1, j, diff - 1)
            return False

        return DFS(0, 0, 0)        
    
    

    @cache
    def get_number(self, s, i):
        """
        Returns Tuple[number_string, s_index_after_number]
        Example:
        - input: i18n, 1     output: "18", 3
        """
        number = ''
        left = right = i
        while right < len(s) and s[right].isdigit():
            right += 1
        return s[left:right], right
        

    @cache
    def get_all_combination(self, num_s):  # Recursion
        """
        Returns possible skip_counts given number string
        Example:
        - input: 123   output: {1+2+3, 1+23, 12+3, 123} = {5, 24, 15, 123}
        """
        ans = {int(num_s)}
        for i in range(1, len(num_s)): 
            for y in self.get_all_combination(num_s[i:]):
                for x in self.get_all_combination(num_s[:i]):
                    ans.add(x + y)  # ans |= {x + y}
        return ans

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