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2303-unique-substrings-with-equal-digit-frequency/README.md

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+Given a digit string <code>s</code>, return <em>the number of <strong>unique substrings </strong>of </em><code>s</code><em> where every digit appears the same number of times.</em>
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;1212&quot;
+<strong>Output:</strong> 5
+<strong>Explanation:</strong> The substrings that meet the requirements are &quot;1&quot;, &quot;2&quot;, &quot;12&quot;, &quot;21&quot;, &quot;1212&quot;.
+Note that although the substring &quot;12&quot; appears twice, it is only counted once.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;12321&quot;
+<strong>Output:</strong> 9
+<strong>Explanation:</strong> The substrings that meet the requirements are &quot;1&quot;, &quot;2&quot;, &quot;3&quot;, &quot;12&quot;, &quot;23&quot;, &quot;32&quot;, &quot;21&quot;, &quot;123&quot;, &quot;321&quot;.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 1000</code></li>
+	<li><code>s</code> consists of digits.</li>
+</ul>

2303-unique-substrings-with-equal-digit-frequency/solution.py

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+# Approach 1: Optimized Brute Force
+
+# Time: O(n^3)
+# Space: O(n^3)
+
+class Solution:
+    def equalDigitFrequency(self, s: str) -> int:
+        n = len(s)
+        valid_substrings = set()
+
+        for start in range(n):
+            digit_freq = [0] * 10 # Frequency array for digits 0-9
+
+            for end in range(start, n):
+                digit_freq[ord(s[end]) - ord('0')] += 1
+
+                # Variable to store the frequency all digits must match
+                common_freq = 0
+                is_valid = True
+
+                for count in digit_freq:
+                    if count == 0:
+                        continue # Skip digits not in the substring
+                    if common_freq == 0:
+                        # First digit found, set common_frequency
+                        common_freq = count
+                    if common_freq != count:
+                        # Mismatch in frequency, mark as invalid
+                        is_valid = False
+                        break
+
+                if is_valid:
+                    substring = s[start : end + 1]
+                    valid_substrings.add(substring)
+
+        return len(valid_substrings)
+