<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>


# TC=O(3^N).
# SC=O(2^N).
# `temp_list`` can be used for debuging the result of palindrome.
class Solution:
    def checkPartitioning(self, s: str) -> bool:
        # def helper(start: int, temp_list: list, remain: int):
        def helper(start: int, remain: int):
            if start > N:
                return False
            if remain == 1:
                return count[start][N]
            for end in range(start, N+1):
                if count[start][end]: 
                    # temp_list.append(s[start:end+1])
                    # if helper(end+1, temp_list, remain - 1):
                    if helper(end+1, remain-1):
                        return True
                    # temp_list.pop()
            return False
        N = len(s)
        s = '#' + s
        count = [[0] * (N+1) for _ in range(N+1)]
        for i in range(1, N+1):
            count[i][i] = 1
        for length in range(2, N+1):
            for i in range(1, N + 1 - length + 1):
                j = i + length - 1
                if s[i] == s[j] and (j - i <= 2 or count[i+1][j-1]):
                    count[i][j] = 1
        # return helper(1, list(), 3)
        return helper(1, 3)
                
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