<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:
        # Utility function to check if a character is a digit
        def is_digit(char: str) -> bool:
            return char.isdigit()

        length_s1 = len(s1)
        length_s2 = len(s2)
        # Initialize a 2D list of sets to use as a dynamic programming table
        dp = [[set() for _ in range(length_s2 + 1)] for _ in range(length_s1 + 1)]

        # Initialize the first cell of the dynamic programming table
        dp[0][0].add(0)

        # Iterate through all positions in both strings
        for i in range(length_s1 + 1):
            for j in range(length_s2 + 1):
                for delta in dp[i][j]:
                    # When s1 contains a number
                    number = 0
                    if delta <= 0:
                        for p in range(i, min(i + 3, length_s1)):
                            if s1[p].isdigit():
                                number = number * 10 + int(s1[p])
                                dp[p + 1][j].add(delta + number)
                            else:
                                # If it's not a digit, stop the sequence
                                break
                
                    # When s2 contains a number
                    number = 0
                    if delta >= 0:
                        for q in range(j, min(j + 3, length_s2)):
                            if s2[q].isdigit():
                                number = number * 10 + int(s2[q])
                                dp[i][q + 1].add(delta - number)
                            else:
                                # If it's not a digit, stop the sequence
                                break
                
                    # Assuming that the numbers might represent a single character
                    # in either s1 or s2 if the next char is a letter
                    if i < length_s1 and delta < 0 and not s1[i].isdigit():
                        dp[i + 1][j].add(delta + 1)
                    if j < length_s2 and delta > 0 and not s2[j].isdigit():
                        dp[i][j + 1].add(delta - 1)
                
                    # Check if current characters in both strings match
                    # and have equal length (delta is zero)
                    if i < length_s1 and j < length_s2 and delta == 0 and s1[i] == s2[j]:
                        dp[i + 1][j + 1].add(0)

        # Return true if a match is found at the end of both strings
        # with no remaining length difference
        return 0 in dp[length_s1][length_s2]
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