<p>Given an array of string <code>words</code>, return <em>all strings in </em><code>words</code><em> that is a <strong>substring</strong> of another word</em>. You can return the answer in <strong>any order</strong>.</p>

<p>A <strong>substring</strong> is a contiguous sequence of characters within a string</p>

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

<pre>
<strong>Input:</strong> words = [&quot;mass&quot;,&quot;as&quot;,&quot;hero&quot;,&quot;superhero&quot;]
<strong>Output:</strong> [&quot;as&quot;,&quot;hero&quot;]
<strong>Explanation:</strong> &quot;as&quot; is substring of &quot;mass&quot; and &quot;hero&quot; is substring of &quot;superhero&quot;.
[&quot;hero&quot;,&quot;as&quot;] is also a valid answer.
</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;leetcode&quot;,&quot;et&quot;,&quot;code&quot;]
<strong>Output:</strong> [&quot;et&quot;,&quot;code&quot;]
<strong>Explanation:</strong> &quot;et&quot;, &quot;code&quot; are substring of &quot;leetcode&quot;.
</pre>

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

<pre>
<strong>Input:</strong> words = [&quot;blue&quot;,&quot;green&quot;,&quot;bu&quot;]
<strong>Output:</strong> []
<strong>Explanation:</strong> No string of words is substring of another string.
</pre>

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

<ul>
	<li><code>1 &lt;= words.length &lt;= 100</code></li>
	<li><code>1 &lt;= words[i].length &lt;= 30</code></li>
	<li><code>words[i]</code> contains only lowercase English letters.</li>
	<li>All the strings of <code>words</code> are <strong>unique</strong>.</li>
</ul>

class Solution {
public:
    vector<string> stringMatching(vector<string>& words) {
        vector<string> result;
        unordered_set<string> unique;
        
        for(const string& word: words) {
            const int n = word.size();
            // build lps
            int i = 1, prevLPS = 0;
            vector<int> lps(n, 0);

            while(i < n) {
                if(word[i] == word[prevLPS]) {
                    lps[i++] = ++prevLPS;
                }
                else if(prevLPS == 0) {
                    lps[i++] = 0;
                }
                else {
                    prevLPS = lps[prevLPS - 1];
                }
            }
            // finish

            for(const string& word2: words) {
                if(word2 != word) {
                    const int m = word2.size();
                    int j = 0;
                    i = 0;

                    while(i < m) {
                        if(word2[i] == word[j]) {
                            ++i;
                            ++j;
                        }
                        else if(j == 0) {
                            ++i;
                        }
                        else {
                            j = lps[j - 1];
                        }

                        if(j == n) {
                            unique.insert(word);
                            break;
                        }
                    }
                }
            }
        }

        for(const string& s: unique) {
            result.push_back(s);
        }

        return result;
    }
};
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