Let <code>dp[i]</code> represent the length of the longest subsequence ending with <code>words[i]</code> that satisfies the conditions.

<code>dp[i] =</code> (maximum value of <code>dp[j]</code>) <code>+ 1</code> for indices <code>j < i</code>, where <code>groups[i] != groups[j]</code>, <code>words[i]</code> and <code>words[j]</code> are equal in length, and the hamming distance between <code>words[i]</code> and <code>words[j]</code> is exactly <code>1</code>.

Keep track of the <code>j</code> values used to achieve the maximum <code>dp[i]</code> for each index <code>i</code>.

The expected array's length is <code>max(dp[0:n])</code>, and starting from the index having the maximum value in <code>dp</code>, we can trace backward to get the words.