<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 lcs(string s,string t) {
        vector<int> prev(s.length()+1,0);
        for(int i=0;i<t.length();i++){
            vector<int> curr(s.length()+1);
            curr[0]=0;
            for(int j=1;j<=s.length();j++){
                if(t[i]==s[j-1]) curr[j]=prev[j-1]+1;
                else curr[j]=max(curr[j-1],prev[j]);
            }
            prev = curr;
        }
        return prev[s.length()];
    }
    int findLUSlength(vector<string>& strs) {
        int ans = 0;
        for(int i=0;i<strs.size();i++){
            bool flag = 1;
            for(int j=0;j<strs.size();j++){
                if(i!=j and lcs(strs[i],strs[j])==strs[i].length()){
                    flag=0;
                    break;
                } 
            }
            if(flag) ans = max(ans, (int)strs[i].length());
        }
        if(ans) return ans;
        else return -1;
    }
};
14641