<p>Given a string <code>s</code>, partition <code>s</code> such that every <span data-keyword="substring-nonempty">substring</span> of the partition is a <span data-keyword="palindrome-string">palindrome</span>.</p>

<p>Return <em>the <strong>minimum</strong> cuts needed for a palindrome partitioning of</em> <code>s</code>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;aab&quot;
<strong>Output:</strong> 1
<strong>Explanation:</strong> The palindrome partitioning [&quot;aa&quot;,&quot;b&quot;] could be produced using 1 cut.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;a&quot;
<strong>Output:</strong> 0
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;ab&quot;
<strong>Output:</strong> 1
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 2000</code></li>
	<li><code>s</code> consists of lowercase English letters only.</li>
</ul>

class Solution {
public:
    int n;

    bool isPalindrome(int i, int j, string &s, vector<vector<int>> &memo)
    {
        if(memo[i][j] != -1)
        return memo[i][j];

        if(i == j)
        return memo[i][j] = true;

        while(i < j)
        {
            if(s[i] == s[j])
            {
                ++i;
                --j;
            }

            else
            return memo[i][j] = false;
        }

        return memo[i][j] = true;
    }

    int solve(int i, string &s, vector<int> &dp, vector<vector<int>> &memo)
    {
        if(i == n)
        return 0;

        if(dp[i] != -1)
        return dp[i];

        int mini = INT_MAX;
        for(int j=i; j<n; j++)
        {
            // left substring is inclusive of j
            if(isPalindrome(i, j, s, memo))
            {
                if(j == n - 1)
                mini = min(mini, solve(j + 1, s, dp, memo));

                else
                mini = min(mini, 1 + solve(j + 1, s, dp, memo));
            }
        }

        return dp[i] = mini;
    }

    int minCut(string s) 
    {
        n = s.length();
        // front partition method
        // we check if the left substring is a palindrome.
        // if yes, only then we proceed with right substring.
        // In the worst case, every left substring will of length 1, so n - 1 partitions

        vector<int> dp(n + 1, -1);
        // also memoizing palindrome checker to figure out if the substring has already been checked for
        vector<vector<int>> memo(n, vector<int> (n, -1));

        return solve(0, s, dp, memo);
    }
};
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