<p>There is a strange printer with the following two special properties:</p>

<ul>
	<li>The printer can only print a sequence of <strong>the same character</strong> each time.</li>
	<li>At each turn, the printer can print new characters starting from and ending at any place and will cover the original existing characters.</li>
</ul>

<p>Given a string <code>s</code>, return <em>the minimum number of turns the printer needed to print it</em>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;aaabbb&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> Print &quot;aaa&quot; first and then print &quot;bbb&quot;.
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;aba&quot;
<strong>Output:</strong> 2
<strong>Explanation:</strong> Print &quot;aaa&quot; first and then print &quot;b&quot; from the second place of the string, which will cover the existing character &#39;a&#39;.
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 100</code></li>
	<li><code>s</code> consists of lowercase English letters.</li>
</ul>

class Solution {
public:
    string s;
    int dp[105][105][200];
    int sol(int st , int nd, char last = '#'){
        if(st > nd){
            return 0;
        }
        if(st == nd){
            return s[st] == last ? 0 : 1;
        }
        if(dp[st][nd][last]!=-1){
            return dp[st][nd][last];
        }
        int ans = INT_MAX;
        for(int i = st; i+1<=nd; ++i){
            ans = min(ans, sol(st, i, last) + sol(i+1,nd, last));
        }
        if(s[st] == s[nd]){
            int cost= 1+sol(st+1,nd-1,s[st]);
            ans = min(ans, cost);
        }
        return dp[st][nd][last] = ans;

    }
    int strangePrinter(string s) {
        this->s = s;
        memset(dp,-1,sizeof(dp));
        return sol(0,s.size()-1);
    }
};
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