<p>Given a string <code>s</code>, you need to partition it into one or more <strong>balanced</strong> <span data-keyword="substring">substrings</span>. For example, if <code>s == &quot;ababcc&quot;</code> then <code>(&quot;abab&quot;, &quot;c&quot;, &quot;c&quot;)</code>, <code>(&quot;ab&quot;, &quot;abc&quot;, &quot;c&quot;)</code>, and <code>(&quot;ababcc&quot;)</code> are all valid partitions, but <code>(&quot;a&quot;, <strong>&quot;bab&quot;</strong>, &quot;cc&quot;)</code>, <code>(<strong>&quot;aba&quot;</strong>, &quot;bc&quot;, &quot;c&quot;)</code>, and <code>(&quot;ab&quot;, <strong>&quot;abcc&quot;</strong>)</code> are not. The unbalanced substrings are bolded.</p>

<p>Return the <strong>minimum</strong> number of substrings that you can partition <code>s</code> into.</p>

<p><strong>Note:</strong> A <strong>balanced</strong> string is a string where each character in the string occurs the same number of times.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;fabccddg&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p>We can partition the string <code>s</code> into 3 substrings in one of the following ways: <code>(&quot;fab, &quot;ccdd&quot;, &quot;g&quot;)</code>, or <code>(&quot;fabc&quot;, &quot;cd&quot;, &quot;dg&quot;)</code>.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">s = &quot;abababaccddb&quot;</span></p>

<p><strong>Output:</strong> <span class="example-io">2</span></p>

<p><strong>Explanation:</strong></p>

<p>We can partition the string <code>s</code> into 2 substrings like so: <code>(&quot;abab&quot;, &quot;abaccddb&quot;)</code>.</p>
</div>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
	<li><code>s</code> consists only of English lowercase letters.</li>
</ul>

class Solution:
    def minimumSubstringsInPartition(self, s: str) -> int:
        @cache
        def best(i=0):
            if i == len(s):
                return 0
            res = float('inf')
            
            counts = [0] * 26
            
            def valid():
                target = max(counts)
                for val in counts:
                    if val != 0 and val != target:
                        return False
                return True
            
            for j in range(i, len(s)):
                counts[ord(s[j]) - ord('a')] += 1
                if valid():
                    res = min(res, 1 + best(j + 1))
            
            return res
        
        return best()
18100