<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 helper(int i, int j, string& s1, string& s2, vector<vector<int>>& dp) {
        if (i >= s1.size() || j >= s2.size())
            return 0;
        if (dp[i][j] != -1)
            return dp[i][j];
        if (s1[i] == s2[j]) {
            return dp[i][j] = 1 + helper(i + 1, j + 1, s1, s2, dp);
        } else {
            return dp[i][j] = max(helper(i + 1, j, s1, s2, dp),
                                  helper(i, j + 1, s1, s2, dp));
        }
    }
    int subsequence(string& s1, string& s2) {
        int m = s1.length();
        int n = s2.length();
        vector<vector<int>> dp(m + 1, vector<int>(n + 1, -1));
        return helper(0, 0, s1, s2, dp);
    }
    int findLUSlength(vector<string>& strs) {
        int ans = -1;

        for (int i = 0; i < strs.size(); i++) {
            int cnt = 1;
            string x = strs[i];
            for (int j = 0; j < strs.size(); j++) {
                if (i != j) {
                    string y = strs[j];
                    int comm = subsequence(x, y)==x.size();
                    cout << comm << " " << x << " " << y << endl;
                    if (comm)
                        {
                            cnt=0;
                            break;
                        }
                }
            }
            if (cnt == 1)
                ans = max(ans, (int)x.length());
        }
        return ans;
    }
};
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