<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]:
        #whats the prev longest up till [i]
        #then find the max amount all start
        #longest subsequence where adjacent indices groups are different and hamming distance == 1
        # then backtrack the subsequence and get the result
        #take/not take with prev
        #if meets criteria we can take, or we can skip
        #take (i+1,i)
        #skip (i+1,) 
        def get_hamming(w1,w2):
            h = 0
            for i in range(len(w1)):
                if w1[i] != w2[i]:
                    h +=1
            return h

        prev = [i for i in range(len(words))]
        dp = [1]*len(words)
        for i in range(len(words)):
            for j in range(i):
                if groups[i] != groups[j] and len(words[i]) == len(words[j]) and get_hamming(words[i],words[j]) <= 1:
                    length = 1 + dp[j]
                    if length > dp[i]:
                        dp[i] = length
                        prev[i] = j
        
        end = max(range(len(words)),key=lambda i: dp[i])

        res = [words[end]]
        while end != prev[end]:
            end = prev[end]
            res.append(words[end])
        return res[::-1]

            
        def dfs(i,prev):
            # if prev == 2:
            #     print('')
            if i == len(words):
                return 0
            key = (i,prev)
            if key in cache:
                return cache[key]
            
            longest = dfs(i+1,prev)

            if prev == -1 or (groups[i] != groups[prev] and len(words[prev]) == len(words[i]) and get_hamming(words[prev],words[i]) <= 1):
                length = 1 + dfs(i+1,i)
                if length > longest:
                    par[prev] = i
                    longest = length

            cache[key] = longest
            return longest

        par = {i:i for i in range(len(words))} #CREATE THE NEXT ARRAY SET AS WE GO ALONG
        cache = {}
        dfs(0,-1)

        i,_ = max(cache,key = cache.get)
        res = [words[i]]
        while i != len(words) and i != par[i]:
            i = par[i]
            res.append(words[i])
        
        return res

        
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