Sync LeetCode submission Runtime - 1787 ms (65.90%), Memory - 25.4 MB (9.25%)

Author: jimit105

3569-count-of-substrings-containing-every-vowel-and-k-consonants-ii/README.md

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+<p>You are given a string <code>word</code> and a <strong>non-negative</strong> integer <code>k</code>.</p>
+
+<p>Return the total number of <span data-keyword="substring-nonempty">substrings</span> of <code>word</code> that contain every vowel (<code>&#39;a&#39;</code>, <code>&#39;e&#39;</code>, <code>&#39;i&#39;</code>, <code>&#39;o&#39;</code>, and <code>&#39;u&#39;</code>) <strong>at least</strong> once and <strong>exactly</strong> <code>k</code> consonants.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">word = &quot;aeioqq&quot;, k = 1</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">0</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>There is no substring with every vowel.</p>
+</div>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">word = &quot;aeiou&quot;, k = 0</span></p>
+
+<p><strong>Output:</strong> <span class="example-io">1</span></p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The only substring with every vowel and zero consonants is <code>word[0..4]</code>, which is <code>&quot;aeiou&quot;</code>.</p>
+</div>
+
+<p><strong class="example">Example 3:</strong></p>
+
+<div class="example-block">
+<p><strong>Input:</strong> <span class="example-io">word = &quot;</span>ieaouqqieaouqq<span class="example-io">&quot;, k = 1</span></p>
+
+<p><strong>Output:</strong> 3</p>
+
+<p><strong>Explanation:</strong></p>
+
+<p>The substrings with every vowel and one consonant are:</p>
+
+<ul>
+	<li><code>word[0..5]</code>, which is <code>&quot;ieaouq&quot;</code>.</li>
+	<li><code>word[6..11]</code>, which is <code>&quot;qieaou&quot;</code>.</li>
+	<li><code>word[7..12]</code>, which is <code>&quot;ieaouq&quot;</code>.</li>
+</ul>
+</div>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>5 &lt;= word.length &lt;= 2 * 10<sup>5</sup></code></li>
+	<li><code>word</code> consists only of lowercase English letters.</li>
+	<li><code>0 &lt;= k &lt;= word.length - 5</code></li>
+</ul>

3569-count-of-substrings-containing-every-vowel-and-k-consonants-ii/solution.py

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+# Approach 1: Sliding Window
+
+# Time: O(n)
+# Space: O(n)
+
+class Solution:
+    def _is_vowel(self, c):
+        return c in ['a', 'e', 'i', 'o', 'u']
+    
+    def countOfSubstrings(self, word: str, k: int) -> int:
+        num_valid_substrings = 0
+        start = end = 0
+        vowel_count = {}
+        consonant_count = 0
+        next_consonant = [0] * len(word) # Array to compute index of next consonant for all indices
+        next_consonant_index = len(word)
+
+        for i in range(len(word) - 1, -1, -1):
+            next_consonant[i] = next_consonant_index
+            if not self._is_vowel(word[i]):
+                next_consonant_index = i
+
+        while end < len(word):
+            new_letter = word[end]
+            if self._is_vowel(new_letter):
+                vowel_count[new_letter] = vowel_count.get(new_letter, 0) + 1
+            else:
+                consonant_count += 1
+
+            while consonant_count > k:
+                # Shrink window
+                start_letter = word[start]
+                if self._is_vowel(start_letter):
+                    vowel_count[start_letter] -= 1
+                    if vowel_count[start_letter] == 0:
+                        del vowel_count[start_letter]
+                else:
+                    consonant_count -= 1
+                start += 1
+
+            while start < len(word) and len(vowel_count) == 5 and consonant_count == k:
+                # Try to shrink if window is valid
+                num_valid_substrings += next_consonant[end] - end
+                start_letter = word[start]
+                if self._is_vowel(start_letter):
+                    vowel_count[start_letter] -= 1
+                    if vowel_count[start_letter] == 0:
+                        del vowel_count[start_letter]
+                else:
+                    consonant_count -= 1
+
+                start += 1
+            
+            end += 1
+
+        return num_valid_substrings
+