<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:
    string st;
    int n;
    int dp[2001];
    int isTrue[2003][2003];

    void isPalin(){
        for(int i = 0; i<n; ++i){
            isTrue[i][i] = 1;
        }

        for(int i = 0; i<n-1; ++i){
            if(st[i] == st[i+1]){
                isTrue[i][i+1] = 1;
            }
        }
        for(int j = 2; j<n; ++j){
            for(int i = 0; i<j; ++i){
                if(st[i] == st[j] && (j-i == 1 || isTrue[i+1][j-1] == 1)){
                    isTrue[i][j] = 1;
                }
            }
        }
    }

    int rec(int idx){
        if(idx == n){
            return 0;
        }
        if(dp[idx] != -1){
            return dp[idx];
        }
        int ans = 2001;
        string str = "";
        for(int i = idx; i<n; ++i){
            str += st[i];
            if(isTrue[idx][i] == 1){
                ans = min(ans , 1 + rec(i+1));
            }
        }
        return dp[idx] = ans;
    }
    int minCut(string s) {
        st = s;
        n = st.size();
        cout<<n<<endl;
        isPalin();
        memset(dp , -1 , sizeof(dp));
        return rec(0)-1;
    }
};
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