<p>Given a string <code>s</code>, return <code>true</code> <em>if it is possible to split the string</em> <code>s</code> <em>into three <strong>non-empty</strong> palindromic substrings. Otherwise, return </em><code>false</code>.​​​​​</p>

<p>A string is said to be palindrome if it the same string when reversed.</p>

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

<pre>
<strong>Input:</strong> s = &quot;abcbdd&quot;
<strong>Output:</strong> true
<strong>Explanation: </strong>&quot;abcbdd&quot; = &quot;a&quot; + &quot;bcb&quot; + &quot;dd&quot;, and all three substrings are palindromes.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;bcbddxy&quot;
<strong>Output:</strong> false
<strong>Explanation: </strong>s cannot be split into 3 palindromes.
</pre>

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

<ul>
	<li><code>3 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code>​​​​​​ consists only of lowercase English letters.</li>
</ul>

class Solution:
    def checkPartitioning(self, s: str) -> bool:
        # 3個所以要用一個2維的
        # 先寫搜尋好了
        memo = {}
        n = len(s)
        def dfs(i,k):
            if k == 1:
                return s[:i+1]==s[:i+1][::-1]
            if (i,k) in memo:
                return memo[(i,k)]
            
            for j in range(i):
                if dfs(j,k-1) and s[j+1:i+1] == s[j+1:i+1][::-1]:
                    memo[(i,k)]=True
                    return True
            memo[(i,k)] = False
            return False
        ret = dfs(n-1,3)
        #print(memo)
        return ret
17800