<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 {
public:

    int dp1[2001][2001];
    int dp[2001][2001];
    string s;
    int n;
    int palindromecheck(int i, int j) {
        if (i == j)
            return 0;
        if (i == j - 1)
            return (s[i] == s[j]) ? 0 : 1;
        auto& x = dp1[i][j];
        if (x != -1) {
            return x;
        }
        return x = (palindromecheck(i + 1, j - 1) + ((s[i] == s[j]) ? 0 : 1));
    }
    int f(int indx, int k) {
        if (indx >= n) {
            return k == 0 ? 0 : 1e9;
        }
        if (k < 0) {
            return 1e9;
        }
        int mini = 1e9;
        auto& x = dp[indx][k];
        if (x != -1)
            return x;
        for (int i = indx; i < n; i++) {
            mini = min(mini, palindromecheck(indx, i) + f(i + 1, k - 1));
        }
        return x = mini;
    }
    bool checkPartitioning(string s) {
        this->s = s;
        memset(dp1,-1,sizeof(dp1));
        memset(dp,-1,sizeof(dp));
        n = s.size();
        return (f(0,3)==0);
    }
};
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