<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:
    unordered_map<int, int> mp; // Memoization with bitmask as the key

    // Recursive function to find the maximum length
    int func(int ind, vector<string>& arr, int taken) {
        if (ind == arr.size()) return 0; // Base case: end of the array

        if (mp.find(taken) != mp.end()) return mp[taken]; // Memoization check

        string curr = arr[ind];
        int currMask = 0; // Bitmask for the current string
        bool isUnique = true;

        // Create a bitmask for the current string and check for duplicates
        for (char c : curr) {
            int bit = c - 'a';
            if (currMask & (1 << bit)) { // Duplicate character in the current string
                isUnique = false;
                break;
            }
            currMask |= (1 << bit); // Set the bit corresponding to the character
        }

        // Check if the current string has unique characters and doesn't overlap with 'taken'
        int take = 0;
        if (isUnique && (taken & currMask) == 0) {
            // Include the current string
            take = curr.size() + func(ind + 1, arr, taken | currMask); 
        }

        // Exclude the current string
        int nt = func(ind + 1, arr, taken); 

        // Store and return the maximum length
        return mp[taken] = max(take, nt);
    }

    int maxLength(vector<string>& arr) {
        return func(0, arr, 0); // Start recursion with index 0 and an empty bitmask
    }
};

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