Sync LeetCode submission Runtime - 368 ms (7.91%), Memory - 17.9 MB (39.71%)

Author: jimit105

1460-number-of-substrings-containing-all-three-characters/README.md

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+<p>Given a string <code>s</code>&nbsp;consisting only of characters <em>a</em>, <em>b</em> and <em>c</em>.</p>
+
+<p>Return the number of substrings containing <b>at least</b>&nbsp;one occurrence of all these characters <em>a</em>, <em>b</em> and <em>c</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;abcabc&quot;
+<strong>Output:</strong> 10
+<strong>Explanation:</strong> The substrings containing&nbsp;at least&nbsp;one occurrence of the characters&nbsp;<em>a</em>,&nbsp;<em>b</em>&nbsp;and&nbsp;<em>c are &quot;</em>abc<em>&quot;, &quot;</em>abca<em>&quot;, &quot;</em>abcab<em>&quot;, &quot;</em>abcabc<em>&quot;, &quot;</em>bca<em>&quot;, &quot;</em>bcab<em>&quot;, &quot;</em>bcabc<em>&quot;, &quot;</em>cab<em>&quot;, &quot;</em>cabc<em>&quot; </em>and<em> &quot;</em>abc<em>&quot; </em>(<strong>again</strong>)<em>. </em>
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;aaacb&quot;
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> The substrings containing&nbsp;at least&nbsp;one occurrence of the characters&nbsp;<em>a</em>,&nbsp;<em>b</em>&nbsp;and&nbsp;<em>c are &quot;</em>aaacb<em>&quot;, &quot;</em>aacb<em>&quot; </em>and<em> &quot;</em>acb<em>&quot;.</em><em> </em>
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;abc&quot;
+<strong>Output:</strong> 1
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>3 &lt;= s.length &lt;= 5 x 10^4</code></li>
+	<li><code>s</code>&nbsp;only consists of&nbsp;<em>a</em>, <em>b</em> or <em>c&nbsp;</em>characters.</li>
+</ul>

1460-number-of-substrings-containing-all-three-characters/solution.py

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+# Approach 1: Sliding Window
+
+# Time: O(n)
+# Space: O(1)
+
+class Solution:
+    def numberOfSubstrings(self, s: str) -> int:
+        left = right = total = 0
+        freq = [0] * 3
+
+        while right < len(s):
+            freq[ord(s[right]) - ord('a')] += 1
+
+            while self._has_all_chars(freq):
+                # All substrings from current window to end are valid
+                # Add count of valid substrings
+                total += len(s) - right
+
+                # Remove leftmost character and move left pointer
+                freq[ord(s[left]) - ord('a')] -= 1
+                left += 1
+
+            right += 1
+
+        return total
+
+    def _has_all_chars(self, freq):
+        return all(f > 0 for f in freq)
+
+
+