<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 func(string& s, int l, int r, vector<vector<int>>& dp) {
        // Base cases
        if (l == r) return 1;  // A single character needs one operation
        if (l > r) return 0;    // No operations needed for invalid ranges
        
        // Return precomputed value if already solved
        if (dp[l][r] != -1) return dp[l][r];

        // Optimize consecutive identical characters
        int i = l;
        while (i < r && s[i + 1] == s[i]) {
            i++;
        }

        // Start with the worst case (all characters printed separately after the first group)
        int ans = 1 + func(s, i + 1, r, dp);
        
        // Try to merge other characters with the current group
        for (int j = i + 1; j <= r; j++) {
            if (s[i] == s[j]) {
                // Attempt to minimize by considering parts before and after j
                ans = min(ans, func(s, i + 1, j - 1, dp) + func(s, j, r, dp));
            }
        }

        // Memoize the result
        return dp[l][r] = ans;
    }

    int strangePrinter(string s) {
        int n = s.size();
        if (n == 0) return 0;

        // Initialize the memoization table
        vector<vector<int>> dp(n, vector<int>(n, -1));

        // Compute the result for the entire string
        return func(s, 0, n - 1, dp);
    }
};

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