Minimum Substring Partition of Equal Character Frequency
<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>
Hash Table; String; Dynamic Programming; Counting
class Solution:
    def minimumSubstringsInPartition(self, s: str) -> int:
        @cache
        def dfs(i: int) -> int:
            if i >= n:
                return 0
            cnt = defaultdict(int)
            freq = defaultdict(int)
            ans = n - i
            for j in range(i, n):
                if cnt[s[j]]:
                    freq[cnt[s[j]]] -= 1
                    if not freq[cnt[s[j]]]:
                        freq.pop(cnt[s[j]])
                cnt[s[j]] += 1
                freq[cnt[s[j]]] += 1
                if len(freq) == 1 and (t := 1 + dfs(j + 1)) < ans:
                    ans = t
            return ans

        n = len(s)
        return dfs(0)
