<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:
    void computeLPS(string p,int m,vector<int>&lps){
     int len=0;
     lps[0]=0;
     int i=1;
     while(i<m){
          if(p[i]==p[len]) lps[i++]=++len;
          else{
               if(len!=0) len=lps[len-1];
               else lps[i++]=0;
          }
     }
}

bool kmp(string s,string p){
     int n=s.size();
     int m=p.size();
     vector<int>lps(m);
     computeLPS(p,m,lps);
     int i=0,j=0;
     while((n-i)>=(m-j)){
          if(s[i]==p[j]){
               i++;
               j++;
          }
          if(j==m){
               return 1;
               j=lps[j-1];
          }
          else if(i<n && s[i]!=p[j]){
               if(j!=0) j=lps[j-1];
               else i++;
          }
     }
     return 0;
}
    vector<string> stringMatching(vector<string>& words) {
        sort(words.begin(),words.end(),[](string a,string b){
            return a.size()<b.size();
        });
        vector<string>res;
        int n=words.size();
        for(int i=0;i<n;i++){
            for(int j=i+1;j<n;j++){
                if(kmp(words[j],words[i])){
                    res.push_back(words[i]);
                    break;
                }
            }
        }
        return res;
    }
};
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