<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 solve(int l, int r, string& s,map<pair<int,int>,int>& dp){
        if(l == r)
            return 1;
        else if(l > r)
            return 0;
        if(dp.count({l,r})!=0)
        return dp[{l,r}];
        int i = l+1;
        while(i <= r && s[i] == s[l])
            i++;
        if(i == r+1)
            return 1;
        int normal = 1 + solve(i,r,s,dp);
        int aage_ka = INT_MAX;
        for(int j = i; j <= r; j++){
            if(s[l] == s[j]){
                
                int x = solve(i,j-1,s,dp) + solve(j,r,s,dp);
                
                aage_ka = min(aage_ka, x);
            }
        }

        return dp[{l,r}] = min(aage_ka, normal);
    }

    int strangePrinter(string s) {
        int n = s.length() ;
        map<pair<int,int>,int>dp;
        return solve(0, n-1, s,dp);
    }
};

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