<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:
        def prepare(s):
            res = []
            pos = 0
            while pos < len(s):
                if s[pos].isalpha():
                    res.append(s[pos])
                    pos += 1
                else:
                    end = pos
                    while end < len(s) and s[end].isdigit():
                        end += 1
                    jumps = set()
                    if end == pos + 1:
                        jumps.add(int(s[pos]))
                    elif end == pos + 2:
                        jumps.add(int(s[pos]) + int(s[pos + 1]))
                        jumps.add(int(s[pos: end]))
                    else:
                        assert end == pos + 3
                        jumps.add(int(s[pos]) + int(s[pos + 1]) + int(s[pos + 2]))
                        jumps.add(int(s[pos: end]))
                        jumps.add(int(s[pos]) + int(s[pos + 1: end]))
                        jumps.add(int(s[pos: pos + 2]) + int(s[pos + 2]))
                    res.append(list(jumps))
                    pos = end
            return res
        s1 = prepare(s1)
        s2 = prepare(s2)

        @cache
        def match(a, b, diff):
            if diff == 0:
                if a == len(s1) or b == len(s2):
                    return a == len(s1) and b == len(s2)
                if isinstance(s1[a], str) and isinstance(s2[b], str):
                    if s1[a] != s2[b]:
                        return False
                    return match(a + 1, b + 1, 0)
            if diff <= 0: # b's has more numbers.need to advance a's side
                if a == len(s1):
                    return False
                if isinstance(s1[a], str):
                    return match(a + 1, b, diff + 1)
                else:
                    for d in s1[a]:
                        if match(a + 1, b, diff + d):
                            return True
                    return False
            # diff > 0 a's side has more number.need to advance b's side now
            if b == len(s2):
                return False
            if isinstance(s2[b], str):
                return match(a, b + 1, diff - 1)
            else:
                for d in s2[b]:
                    if match(a, b + 1, diff - d):
                        return True
                return False

        return match(0, 0, 0)        
279