<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 {
public:
	int solve(int index, vector<string> ans, vector<string>& arr) {
		if (index >= arr.size()) {
			return 0;
		}
		ans.push_back(arr[index]);
		unordered_map<char, int> freq;
		for (int i = 0; i < ans.size(); i++) {
			for (int j = 0; j < ans[i].size(); j++) {
				freq[ans[i][j]]++;
			}
		}
		auto foundDup = false;
		for (auto& i : freq) {
			if (i.second > 1) {
				foundDup = true;
				break;
			}
		}
		int opt1 = 0, opt2 = 0;
		if (!foundDup) {
			opt1 = arr[index].size() + solve(index+1, ans, arr);
		}
		ans.pop_back();
		opt2 = solve(index+1, ans, arr);
		return max(opt1, opt2);
	}

	int maxLength(vector<string>& arr) {
		vector<string> ans;
		return solve(0, ans, arr);
	}
};
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