<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:
    bool isPalindrome(string &s) {
        int i = 0;
        int j = s.size() - 1;
        while (i < j) {
            if (s[i] != s[j])
                return false;
            i++;
            j--;
        }
        return true;
    }
    int solve(int i, string& s, vector<int>& dp) {
        if (i == s.size()) {
            return 0;
        }
        if (dp[i] != -1)
            return dp[i];
        // check for partition at every index
        // maintain a temp string and minCost
        string temp = "";
        int minCost = INT_MAX;
        for (int j = i; j < s.size(); j++) {
            temp += s[j];
            if (isPalindrome(temp)) {
                // only if this is a palindrome we can create the partition
                int cost = 1 + solve(j + 1, s, dp);
                minCost = min(cost, minCost);
            }
        }
        return dp[i] = minCost;
    }
    int minCut(string s) {
        // if we make n-1 partitions we can have all the chars as pali in itself
        //  but we need minimum

        // This is a front partition problem
        int n = s.size();
        vector<int> dp(n + 1, 0);
        for (int i = n - 1; i >= 0; i--) {
            string temp = "";
            int minCost = INT_MAX;
            for (int j = i; j < s.size(); j++) {
                temp += s[j];
                if (isPalindrome(temp)) {
                    // only if this is a palindrome we can create the partition
                    int cost = 1 + solve(j + 1, s, dp);
                    minCost = min(cost, minCost);
                }
            }
            dp[i] = minCost;
        }
        // The -1 is bcoz we do a partition at the end of the string
        return dp[0] - 1;
    }
};
31395