<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:
        n = len(s)
        sums = defaultdict(lambda: [0] * (n + 1))
        DP = [(i + 1) % (n + 1) for i in range(n + 1)]
        OK = [[True for _ in range(n)] for _ in range(n)]
        for i in range(n):
            for letter in ascii_lowercase:
                sums[letter][i + 1] = sums[letter][i] + (s[i] == letter)
        for j in range(n):
            for i in range(j + 1):
                freq = sums[s[i]][j + 1] - sums[s[i]][i]
                for letter in ascii_lowercase:
                    currFreq = sums[letter][j + 1] - sums[letter][i]
                    if currFreq != 0 and currFreq != freq:
                        OK[i][j] = False
                        break
                if OK[i][j]: DP[j] = min(DP[j], DP[i - 1] + 1)
        return DP[n - 1]
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