<p>Given an array of strings <code>strs</code>, return <em>the length of the <strong>longest uncommon subsequence</strong> between them</em>. If the longest uncommon subsequence does not exist, return <code>-1</code>.</p>

<p>An <strong>uncommon subsequence</strong> between an array of strings is a string that is a <strong>subsequence of one string but not the others</strong>.</p>

<p>A <strong>subsequence</strong> of a string <code>s</code> is a string that can be obtained after deleting any number of characters from <code>s</code>.</p>

<ul>
	<li>For example, <code>&quot;abc&quot;</code> is a subsequence of <code>&quot;aebdc&quot;</code> because you can delete the underlined characters in <code>&quot;a<u>e</u>b<u>d</u>c&quot;</code> to get <code>&quot;abc&quot;</code>. Other subsequences of <code>&quot;aebdc&quot;</code> include <code>&quot;aebdc&quot;</code>, <code>&quot;aeb&quot;</code>, and <code>&quot;&quot;</code> (empty string).</li>
</ul>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> strs = ["aba","cdc","eae"]
<strong>Output:</strong> 3
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> strs = ["aaa","aaa","aa"]
<strong>Output:</strong> -1
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= strs.length &lt;= 50</code></li>
	<li><code>1 &lt;= strs[i].length &lt;= 10</code></li>
	<li><code>strs[i]</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:

int recursion(int i,int j, string &s, string &t, vector<vector<int>> &dp){

    if(i>=s.length() || j>=t.length()) return 0;
    if(dp[i][j] != -1) return dp[i][j];

    if(s[i]==t[j]) return dp[i][j]=1+recursion(i+1,j+1,s,t,dp);

        return dp[i][j] = max(recursion(i+1,j,s,t,dp) , recursion(i,j+1,s,t,dp));
    
}
    int findLUSlength(vector<string>& strs) {
        
        int ans=-1;
        bool flag=true;

        for(int i=0;i<strs.size();i++){
            for(int j=0;j<strs.size();j++){

                if(i != j) {
                
                int length=0;

                vector<vector<int>> dp(strs[i].length(),vector<int>(strs[j].length(),-1));
                length = recursion(0,0,strs[i],strs[j],dp);

              if(length == strs[i].length()) {
                        flag=false;
                        break;
                    }
                }
            }

            if(flag){
                int n=strs[i].length();
               ans=max(ans,n);
            } 
            flag=true;
        }

        return ans;
    }
};
16738