Maximum Length of a Concatenated String with Unique Characters
<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>
Bit Manipulation; Array; String; Backtracking
class Solution {
public:
    int maxLength(vector<string>& arr) {
        vector<int> s = {0};
        int ans = 0;
        for (const string& t : arr) {
            int x = 0;
            for (char c : t) {
                int b = c - 'a';
                if ((x >> b & 1) == 1) {
                    x = 0;
                    break;
                }
                x |= 1 << b;
            }
            if (x > 0) {
                for (int i = s.size() - 1; i >= 0; --i) {
                    int y = s[i];
                    if ((x & y) == 0) {
                        s.push_back(x | y);
                        ans = max(ans, __builtin_popcount(x | y));
                    }
                }
            }
        }
        return ans;
    }
};
