<p>You are given a <strong>0-indexed</strong> array of <strong>unique</strong> strings <code>words</code>.</p>

<p>A <strong>palindrome pair</strong> is a pair of integers <code>(i, j)</code> such that:</p>

<ul>
	<li><code>0 &lt;= i, j &lt; words.length</code>,</li>
	<li><code>i != j</code>, and</li>
	<li><code>words[i] + words[j]</code> (the concatenation of the two strings) is a <span data-keyword="palindrome-string">palindrome</span>.</li>
</ul>

<p>Return <em>an array of all the <strong>palindrome pairs</strong> of </em><code>words</code>.</p>

<p>You must write an algorithm with&nbsp;<code>O(sum of words[i].length)</code>&nbsp;runtime complexity.</p>

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

<pre>
<strong>Input:</strong> words = [&quot;abcd&quot;,&quot;dcba&quot;,&quot;lls&quot;,&quot;s&quot;,&quot;sssll&quot;]
<strong>Output:</strong> [[0,1],[1,0],[3,2],[2,4]]
<strong>Explanation:</strong> The palindromes are [&quot;abcddcba&quot;,&quot;dcbaabcd&quot;,&quot;slls&quot;,&quot;llssssll&quot;]
</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;bat&quot;,&quot;tab&quot;,&quot;cat&quot;]
<strong>Output:</strong> [[0,1],[1,0]]
<strong>Explanation:</strong> The palindromes are [&quot;battab&quot;,&quot;tabbat&quot;]
</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;a&quot;,&quot;&quot;]
<strong>Output:</strong> [[0,1],[1,0]]
<strong>Explanation:</strong> The palindromes are [&quot;a&quot;,&quot;a&quot;]
</pre>

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

<ul>
	<li><code>1 &lt;= words.length &lt;= 5000</code></li>
	<li><code>0 &lt;= words[i].length &lt;= 300</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    bool isPalindrome(string s, int i, int j) {
        if (i == j || i > j) return true;
        if (s[i] != s[j]) return false;
        return isPalindrome(s, i + 1, j - 1);
    }

    vector<vector<int>> palindromePairs(vector<string>& words) {
        set<vector<int> > ans;
        unordered_map<string, bool> strs, strs2;
        unordered_map<string, int> strToIndex, strToIndex2;

        for (int i=0; i<words.size(); i++) {
            string word = words[i];
            strs[word] = true;
            strToIndex[word] = i;
            reverse(word.begin(), word.end());
            strs2[word] = true;
            strToIndex2[word] = i;
        }

        for (int i=0; i<words.size(); i++) {
            // check if can add to ans
            string word = words[i];
            string s = "";

            // handles case one: add word to right of another word, must traverse in reverse
            for (int j=word.size()-1; j>=0; j--) {
                s += word[j];
                if (strs.find(s) != strs.end() && strToIndex[s] != i && isPalindrome(word, 0, j - 1)) {
                    vector<int> tmp {strToIndex[s], i};
                    ans.insert(tmp);
                }
            }

            if (strs.find("") != strs.end() && word != "" && isPalindrome(word, 0, word.size()-1)) {
                vector<int> tmp {strToIndex[""], i};
                vector<int> tmp2 {i, strToIndex[""]};
                ans.insert(tmp);
                ans.insert(tmp2);
            }

            // handles case two: add word to left of another word, must traverse normally
            s = "";
            for (int j=0; j<word.size(); j++) {
                s += word[j];
                if (strs2.find(s) != strs2.end() && strToIndex2[s] != i && isPalindrome(word, j + 1, word.size()-1)) {
                    vector<int> tmp {i, strToIndex2[s]};
                    ans.insert(tmp);
                }
            }
        }
        vector<vector<int> > output;
        for (auto it=ans.begin(); it!=ans.end(); it++) {
            vector<int> tmp = *it;
            output.push_back(tmp);
        }
        return output;
    }
};
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