<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:
        def parse(s):
            res = []
            i = 0
            while i < len(s):
                if s[i].isalpha():
                    res.append(s[i])
                    i += 1
                    continue
                if s[i].isdigit():
                    curr = []
                    while i < len(s) and s[i].isdigit():
                        curr.append(s[i])
                        i += 1
                    res.append(curr)
                    continue
            return res
        
        def getNum(t):
            d = int(''.join(t))
            if len(t) == 1:
                return {d}
            if len(t) == 2:
                a = d // 10
                b = d % 10
                return {a + b, d}
            if len(t) == 3:
                a = d // 100
                b = d % 100 // 10
                c = d % 10
                return {a * 10 + b + c, a + b * 10 + c, a + b + c, d}
        
        def dfs(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
            hashcode = str(i) + str(num1) + str(j) + str(num2)
            if hashcode in self.visited:
                return False
            if i < len(t1) and t1[i][0].isdigit():
                nums = getNum(t1[i])
                for n in nums:
                    if dfs(t1, i + 1, num1 + n, t2, j, num2):
                        return True
                self.visited.add(hashcode)
                return False
            elif j < len(t2) and t2[j][0].isdigit():
                nums = getNum(t2[j])
                for n in nums:
                    if dfs(t1, i, num1, t2, j + 1, num2 + n):
                        return True
                self.visited.add(hashcode)
                return False
            if num1 != 0 and num2 != 0:
                common = min(num1, num2)
                return dfs(t1, i, num1 - common, t2, j, num2 - common)
            elif num1 != 0 and num2 == 0:
                return dfs(t1, i, num1 - 1, t2, j + 1, 0)
            elif num1 == 0 and num2 != 0:
                return dfs(t1, i + 1, 0, t2, j, num2 - 1)
            else:
                if t1[i] != t2[j]:
                    self.visited.add(hashcode)
                    return False
                return dfs(t1, i + 1, 0, t2, j + 1, 0)
        t1, t2 = parse(s1), parse(s2)
        return dfs(t1, 0, 0, t2, 0, 0)
1045