<p>You are given a string array <code>words</code>, and an array <code>groups</code>, both arrays having length <code>n</code>.</p>

<p>The <strong>hamming distance</strong> between two strings of equal length is the number of positions at which the corresponding characters are <strong>different</strong>.</p>

<p>You need to select the <strong>longest</strong> <span data-keyword="subsequence-array">subsequence</span> from an array of indices <code>[0, 1, ..., n - 1]</code>, such that for the subsequence denoted as <code>[i<sub>0</sub>, i<sub>1</sub>, ..., i<sub>k-1</sub>]</code> having length <code>k</code>, the following holds:</p>

<ul>
	<li>For <strong>adjacent</strong> indices in the subsequence, their corresponding groups are <strong>unequal</strong>, i.e., <code>groups[i<sub>j</sub>] != groups[i<sub>j+1</sub>]</code>, for each <code>j</code> where <code>0 &lt; j + 1 &lt; k</code>.</li>
	<li><code>words[i<sub>j</sub>]</code> and <code>words[i<sub>j+1</sub>]</code> are <strong>equal</strong> in length, and the <strong>hamming distance</strong> between them is <code>1</code>, where <code>0 &lt; j + 1 &lt; k</code>, for all indices in the subsequence.</li>
</ul>

<p>Return <em>a string array containing the words corresponding to the indices <strong>(in order)</strong> in the selected subsequence</em>. If there are multiple answers, return <em>any of them</em>.</p>

<p><strong>Note:</strong> strings in <code>words</code> may be <strong>unequal</strong> in length.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">words = [&quot;bab&quot;,&quot;dab&quot;,&quot;cab&quot;], groups = [1,2,2]</span></p>

<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">[&quot;bab&quot;,&quot;cab&quot;]</span></p>

<p><strong>Explanation: </strong>A subsequence that can be selected is <code>[0,2]</code>.</p>

<ul>
	<li><code>groups[0] != groups[2]</code></li>
	<li><code>words[0].length == words[2].length</code>, and the hamming distance between them is 1.</li>
</ul>

<p>So, a valid answer is <code>[words[0],words[2]] = [&quot;bab&quot;,&quot;cab&quot;]</code>.</p>

<p>Another subsequence that can be selected is <code>[0,1]</code>.</p>

<ul>
	<li><code>groups[0] != groups[1]</code></li>
	<li><code>words[0].length == words[1].length</code>, and the hamming distance between them is <code>1</code>.</li>
</ul>

<p>So, another valid answer is <code>[words[0],words[1]] = [&quot;bab&quot;,&quot;dab&quot;]</code>.</p>

<p>It can be shown that the length of the longest subsequence of indices that satisfies the conditions is <code>2</code>.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block" style="border-color: var(--border-tertiary); border-left-width: 2px; color: var(--text-secondary); font-size: .875rem; margin-bottom: 1rem; margin-top: 1rem; overflow: visible; padding-left: 1rem;">
<p><strong>Input: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">words = [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;,&quot;d&quot;], groups = [1,2,3,4]</span></p>

<p><strong>Output: </strong><span class="example-io" style="font-family: Menlo,sans-serif; font-size: 0.85rem;">[&quot;a&quot;,&quot;b&quot;,&quot;c&quot;,&quot;d&quot;]</span></p>

<p><strong>Explanation: </strong>We can select the subsequence <code>[0,1,2,3]</code>.</p>

<p>It satisfies both conditions.</p>

<p>Hence, the answer is <code>[words[0],words[1],words[2],words[3]] = [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;,&quot;d&quot;]</code>.</p>

<p>It has the longest length among all subsequences of indices that satisfy the conditions.</p>

<p>Hence, it is the only answer.</p>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n == words.length == groups.length &lt;= 1000</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
	<li><code>1 &lt;= groups[i] &lt;= n</code></li>
	<li><code>words</code> consists of <strong>distinct</strong> strings.</li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
</ul>

class Solution:
    def getWordsInLongestSubsequence(self, words: List[str], groups: List[int]) -> List[str]:

        def hamming(s1: str, s2: str) -> int:
            if len(s1) != len(s2):
                return 0
            return sum([
                (0 if (c1 == c2) else 1)
                for c1, c2 in zip(s1, s2)
            ])
        
        def allowed(index1: int, index2: int) -> bool:
            return (
                (
                    groups[index1] != groups[index2]
                ) and (
                    hamming(words[index1], words[index2]) == 1
                )
            )
        
        longest_ss_ending_at = [1] * len(words)         # each could be start of new seq
        longest_ss_prior_index = [None] * len(words)    # ... with no prior value

        for j in range(len(words)):
            print(f'[{j=}]')
            currentI = longest_ss_ending_at[j] + 1
            for i in range(j + 1, len(words)):
                # print(f'  [{i=}]')
                if not allowed(j, i):
                    # print(f'    not allowed')
                    continue
                priorI = longest_ss_ending_at[i]
                if currentI > priorI:
                    # print(f'    New best length {currentI} > {priorI}')
                    longest_ss_ending_at[i] = currentI
                    longest_ss_prior_index[i] = j
        answer_length = max(longest_ss_ending_at)
        print(f'{answer_length=}')
        answer_index = [
            index
            for index, length in enumerate(longest_ss_ending_at)
            if length == answer_length
        ]
        print(f'{answer_index=}')
        K = answer_index[0]
        answer = []
        # print(f'Composing answer:')
        while K is not None:
            # print(f'  {K=} {words[K]}')
            answer.append(words[K])
            K = longest_ss_prior_index[K]

        return reversed(answer)

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