<p>You are given an array of strings <code>words</code>. Each element of <code>words</code> consists of <strong>two</strong> lowercase English letters.</p>

<p>Create the <strong>longest possible palindrome</strong> by selecting some elements from <code>words</code> and concatenating them in <strong>any order</strong>. Each element can be selected <strong>at most once</strong>.</p>

<p>Return <em>the <strong>length</strong> of the longest palindrome that you can create</em>. If it is impossible to create any palindrome, return <code>0</code>.</p>

<p>A <strong>palindrome</strong> is a string that reads the same forward and backward.</p>

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

<pre>
<strong>Input:</strong> words = [&quot;lc&quot;,&quot;cl&quot;,&quot;gg&quot;]
<strong>Output:</strong> 6
<strong>Explanation:</strong> One longest palindrome is &quot;lc&quot; + &quot;gg&quot; + &quot;cl&quot; = &quot;lcggcl&quot;, of length 6.
Note that &quot;clgglc&quot; is another longest palindrome that can be created.
</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;ab&quot;,&quot;ty&quot;,&quot;yt&quot;,&quot;lc&quot;,&quot;cl&quot;,&quot;ab&quot;]
<strong>Output:</strong> 8
<strong>Explanation:</strong> One longest palindrome is &quot;ty&quot; + &quot;lc&quot; + &quot;cl&quot; + &quot;yt&quot; = &quot;tylcclyt&quot;, of length 8.
Note that &quot;lcyttycl&quot; is another longest palindrome that can be created.
</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;cc&quot;,&quot;ll&quot;,&quot;xx&quot;]
<strong>Output:</strong> 2
<strong>Explanation:</strong> One longest palindrome is &quot;cc&quot;, of length 2.
Note that &quot;ll&quot; is another longest palindrome that can be created, and so is &quot;xx&quot;.
</pre>

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

<ul>
	<li><code>1 &lt;= words.length &lt;= 10<sup>5</sup></code></li>
	<li><code>words[i].length == 2</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
</ul>

// Solution: Unordered Map
// Run Time: O(n)
// Space: O(n)
// 1. Count frequency of each word in map
// 2. Find opposite word for each word in array
// 3. If opposite word is same as word, then find number of pairs (freq / 2) multiplied by 4 for length
//    and if frequency is odd, record that
// 4. If opposite word is found in map, find number of pairs by taking minimum freq between words
//    and multiply by 4 for length and mark word as done
// 5. If there is an odd frequency for same word from earlier, add 2 to result
class Solution {
public:
    int longestPalindrome(vector<string>& words) {
        unordered_map<string, int> mp;
        int res = 0;
        bool odd = false;

        for (string &w : words) {
            mp[w]++;
        }
        for (auto &i : mp) {
            string w = i.first;
            int freq = i.second;
            string op = string(w.rbegin(), w.rend());

            if (w == op) {
                res += (freq / 2) * 4;
                if (freq % 2 == 1) {
                    odd = true;
                }
            } else if (mp.find(op) != mp.end()) {
                int comp = min(freq, mp[op]);
                res += comp * 4;
                mp[op] = 0;
            }
        }
        if (odd) res += 2;
        return res;
    }
};
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