Sync LeetCode submission Runtime - 11 ms (66.61%), Memory - 17.7 MB (79.05%)

Author: jimit105

1084-find-k-length-substrings-with-no-repeated-characters/README.md

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+<p>Given a string <code>s</code> and an integer <code>k</code>, return <em>the number of substrings in </em><code>s</code><em> of length </em><code>k</code><em> with no repeated characters</em>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;havefunonleetcode&quot;, k = 5
+<strong>Output:</strong> 6
+<strong>Explanation:</strong> There are 6 substrings they are: &#39;havef&#39;,&#39;avefu&#39;,&#39;vefun&#39;,&#39;efuno&#39;,&#39;etcod&#39;,&#39;tcode&#39;.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> s = &quot;home&quot;, k = 5
+<strong>Output:</strong> 0
+<strong>Explanation:</strong> Notice k can be larger than the length of s. In this case, it is not possible to find any substring.
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= s.length &lt;= 10<sup>4</sup></code></li>
+	<li><code>s</code> consists of lowercase English letters.</li>
+	<li><code>1 &lt;= k &lt;= 10<sup>4</sup></code></li>
+</ul>

1084-find-k-length-substrings-with-no-repeated-characters/solution.py

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+# Approach 2: Sliding Window
+
+# n = len(s), m = no. of unique chars (i.e. 26)
+# Time: O(n)
+# Space: O(m) = O(26) = O(1)
+
+class Solution:
+    def numKLenSubstrNoRepeats(self, s: str, k: int) -> int:
+        if k > 26:
+            return 0
+        answer = 0
+        n = len(s)
+
+        left = right = 0
+        freq = [0] * 26
+
+        def get_val(ch):
+            return ord(ch) - ord('a')
+
+        while right < n:
+            freq[get_val(s[right])] += 1
+
+            while freq[get_val(s[right])] > 1:
+                freq[get_val(s[left])] -= 1
+                left += 1
+
+            if right - left + 1 == k:
+                answer += 1
+                
+                freq[get_val(s[left])] -= 1
+                left += 1
+
+            right += 1
+
+        return answer
+
+
+        
+