<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 __init__(self):
        self.visited = set()

    def possiblyEquals(self, s1: str, s2: str) -> bool:
        t1 = self.parse(s1)
        t2 = self.parse(s2)
        return self.dfs(t1, 0, 0, t2, 0, 0)
    
    def parse(self, s):
        t = []
        i = 0
        while i < len(s):
            if s[i].isalpha():
                t.append(s[i])
                i += 1
            else:
                j = i
                while j < len(s) and s[j].isdigit():
                    j += 1
                t.append(s[i:j])
                i = j
        return t
    
    def getNum(self, t):
        d = int(t)
        if len(t) == 1:
            return {d}
        elif len(t) == 2:
            a = d // 10
            b = d % 10
            return {a + b, d}
        else:
            a = d // 100
            b = (d // 10) % 10
            c = d % 10
            return {a + b + c, a + b * 10 + c, a * 10 + b + c, d}
    
    def dfs(self, t1, i, num1, t2, j, num2):
        if i == len(t1) and j == len(t2):
            return num1 == num2
        
        if i == len(t1) and num1 == 0:
            return False
        
        if j == len(t2) and num2 == 0:
            return False
        
        # Hash computation for visited state
        hash_value = i * 1_000_000_000 + num1 * 1_000_000 + j * 1_000 + num2
        if hash_value in self.visited:
            return False
        
        if i < len(t1) and t1[i][0].isdigit():
            nums = self.getNum(t1[i])
            for x in nums:
                if self.dfs(t1, i + 1, num1 + x, t2, j, num2):
                    return True
            self.visited.add(hash_value)
            return False
        
        if j < len(t2) and t2[j][0].isdigit():
            nums = self.getNum(t2[j])
            for x in nums:
                if self.dfs(t1, i, num1, t2, j + 1, num2 + x):
                    return True
            self.visited.add(hash_value)
            return False
        
        if num1 != 0 and num2 != 0:
            common = min(num1, num2)
            return self.dfs(t1, i, num1 - common, t2, j, num2 - common)
        
        if num1 != 0 and num2 == 0:
            return self.dfs(t1, i, num1 - 1, t2, j + 1, 0)
        
        if num1 == 0 and num2 != 0:
            return self.dfs(t1, i + 1, 0, t2, j, num2 - 1)
        
        if t1[i] != t2[j]:
            self.visited.add(hash_value)
            return False
        
        return self.dfs(t1, i + 1, 0, t2, j + 1, 0)


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