<p>You are given an array of strings <code>arr</code>. A string <code>s</code> is formed by the <strong>concatenation</strong> of a <strong>subsequence</strong> of <code>arr</code> that has <strong>unique characters</strong>.</p>

<p>Return <em>the <strong>maximum</strong> possible length</em> of <code>s</code>.</p>

<p>A <strong>subsequence</strong> is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements.</p>

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

<pre>
<strong>Input:</strong> arr = [&quot;un&quot;,&quot;iq&quot;,&quot;ue&quot;]
<strong>Output:</strong> 4
<strong>Explanation:</strong> All the valid concatenations are:
- &quot;&quot;
- &quot;un&quot;
- &quot;iq&quot;
- &quot;ue&quot;
- &quot;uniq&quot; (&quot;un&quot; + &quot;iq&quot;)
- &quot;ique&quot; (&quot;iq&quot; + &quot;ue&quot;)
Maximum length is 4.
</pre>

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

<pre>
<strong>Input:</strong> arr = [&quot;cha&quot;,&quot;r&quot;,&quot;act&quot;,&quot;ers&quot;]
<strong>Output:</strong> 6
<strong>Explanation:</strong> Possible longest valid concatenations are &quot;chaers&quot; (&quot;cha&quot; + &quot;ers&quot;) and &quot;acters&quot; (&quot;act&quot; + &quot;ers&quot;).
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> arr = [&quot;abcdefghijklmnopqrstuvwxyz&quot;]
<strong>Output:</strong> 26
<strong>Explanation:</strong> The only string in arr has all 26 characters.
</pre>

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

<ul>
	<li><code>1 &lt;= arr.length &lt;= 16</code></li>
	<li><code>1 &lt;= arr[i].length &lt;= 26</code></li>
	<li><code>arr[i]</code> contains only lowercase English letters.</li>
</ul>

class Solution:
    def maxLength2(self, arr: List[str]) -> int:
        charSet = set()

        def overlap(charSet, s):
            c = Counter(charSet) + Counter(s)
            return max(c.values()) > 1

        def backtrack(i):
            if i == len(arr):
                return len(charSet)

            res = 0

            if not overlap(charSet, arr[i]):
                for c in arr[i]:
                    charSet.add(c)

                res = backtrack(i + 1)

                for c in arr[i]:
                    charSet.remove(c)

            return max(res, backtrack(i + 1))  # dont concatenate arr[i]

        return backtrack(0)

# Time: O(len(string) ^ n * len(string))
# Space: O(len(string) ^ n)


    def maxLength(self, arr: List[str]) -> int:
        res = 0
        cur = set()

        def backtrack(i):
            nonlocal res
            if i == len(arr):
                res = max(res, len(cur))
                return

            # don't take i
            backtrack(i + 1)

            # take i
            j = 0
            while j < len(arr[i]):
                if arr[i][j] in cur:
                    break
                cur.add(arr[i][j])
                j += 1

            if j == len(arr[i]):
                backtrack(i + 1)

            # backtrack
            j -= 1
            while j >= 0:
                cur.remove(arr[i][j])
                j -= 1

        backtrack(0)
        return res

# Time: O(len(string) ^ n * len(string))
# Space: O(len(string) ^ n)


    def maxLength2(self, arr: List[str]) -> int:
        res = 0

        def backtrack(i, cur):
            nonlocal res
            if i == len(arr):
                res = max(res, len(cur))
                return

            # don't take i
            backtrack(i + 1, cur)

            # take i
            j = 0
            while j < len(arr[i]):
                if arr[i][j] in cur:
                    break
                cur.add(arr[i][j])
                j += 1

            if j == len(arr[i]):
                backtrack(i + 1, cur)

            # backtrack
            j -= 1
            while j >= 0:
                cur.remove(arr[i][j])
                j -= 1

        backtrack(0, set())
        return res

# Time: O(len(string) ^ n * len(string))
# Space: O(len(string) ^ n)
16800