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 {
public:
    int minimumSubstringsInPartition(string s) {
        int n = s.size();
        int f[n];
        memset(f, -1, sizeof(f));
        auto dfs = [&](this auto&& dfs, int i) -> int {
            if (i >= n) {
                return 0;
            }
            if (f[i] != -1) {
                return f[i];
            }
            f[i] = n - i;
            int cnt[26]{};
            unordered_map<int, int> freq;
            for (int j = i; j < n; ++j) {
                int k = s[j] - 'a';
                if (cnt[k]) {
                    freq[cnt[k]]--;
                    if (freq[cnt[k]] == 0) {
                        freq.erase(cnt[k]);
                    }
                }
                ++cnt[k];
                ++freq[cnt[k]];
                if (freq.size() == 1) {
                    f[i] = min(f[i], 1 + dfs(j + 1));
                }
            }
            return f[i];
        };
        return dfs(0);
    }
};
