<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 {
    vector<int> isPalindrome(string& s, bool inPrefix) {
        int n=s.length();
        
        string ss="#";
        for (char ch: s) ss+=ch, ss+='#';
        int m=2*n+1;

        vector<int> rd(m);
        int l=0, r=0;
        for (int i=0; i<m; i++)
        {
            if (i<=r) rd[i] = min(r-i, rd[l+r-i]);
            int ll=i-rd[i]-1, rr=i+rd[i]+1;
            while ((ll>=0 && rr<m) && (ss[ll]==ss[rr])) ll--, rr++, rd[i]++;
            l=i-rd[i], r=i+rd[i];
        }

        vector<int> dp(n);
        for (int i=1; i<m-1; i++)
        {
            l=i-rd[i], r=i+rd[i];
            if (inPrefix && l==0) dp[r/2-1]=1;
            if (!inPrefix && r==m-1) dp[l/2]=1;
        }

        return dp;
    }

public:
    vector<vector<int>> palindromePairs(vector<string>& words) {
        int n = words.size();

        unordered_map<string, int> pmap, smap;
        for (int i=0; i<n; i++)
        {
            string word=words[i]; pmap[word] = i;
            reverse(word.begin(), word.end()); smap[word]=i;
        }

        set<vector<int>> pp;
        for (int i=0; i<n; i++) {
            string w=words[i], suf="", pre="";
            int l=w.length();
            vector<int> preDp=isPalindrome(w, 1), sufDp=isPalindrome(w, 0);
            
            for (int j=l; j>=0; j--) {
                if (j<l) suf+=w[j];
                if (pmap.find(suf) != pmap.end() && pmap[suf] != i && (j==0 || preDp[j-1])) pp.insert({pmap[suf], i});
            }

            for (int j=-1; j<l; j++) {
                if (j>=0) pre+=w[j];
                if (smap.find(pre) != smap.end() && smap[pre]!=i && (j==l-1 || sufDp[j+1])) pp.insert({i, smap[pre]});
            }
        }

        vector<vector<int>> ans;
        for (auto p:pp) ans.push_back(p);
        return ans;
    }
};
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