<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:
   int f(int l,int r,string &s,vector<vector<long long >>&dp){
    if(l>r)return 0;
    if(l==r)return 1;
    if(dp[l][r]!=-1)return dp[l][r];
    int ans = 1+f(l+1,r,s,dp);
    int ans1 = INT_MAX;
    for(int i=l+1; i<=r; i++){
        if(s[i]==s[l]){
            ans1 = min(ans1,f(l,i-1,s,dp)+f(i+1,r,s,dp));
        }
    }
    return dp[l][r] = min(ans,ans1);
   }
    int strangePrinter(string s) {
        int n = s.size();
         vector<vector<long long >>dp(n+1,vector<long long >(n+1,-1));
        return f(0,n-1,s,dp);
    }
};
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