<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
        n = len(s)
        dp = [[False]*3 for _ in range(n)]

        for i in range(n):

            dp[i][0] = s[:i+1]==s[:i+1][::-1]
            for j in range(i):
                if dp[j][0] and s[j+1:i+1][::-1]==s[j+1:i+1]:
                    dp[i][1] = True
                    break
            for j in range(i):
                if dp[j][1] and s[j+1:i+1][::-1]==s[j+1:i+1]:
                    dp[i][2] = True
                    break
                
        
        return dp[-1][2]
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