Sync LeetCode submission Runtime - 19 ms (71.40%), Memory - 49.8 MB (30.11%)

Author: jimit105

2691-count-vowel-strings-in-ranges/README.md

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+<p>You are given a <strong>0-indexed</strong> array of strings <code>words</code> and a 2D array of integers <code>queries</code>.</p>
+
+<p>Each query <code>queries[i] = [l<sub>i</sub>, r<sub>i</sub>]</code> asks us to find the number of strings present in the range <code>l<sub>i</sub></code> to <code>r<sub>i</sub></code> (both <strong>inclusive</strong>) of <code>words</code> that start and end with a vowel.</p>
+
+<p>Return <em>an array </em><code>ans</code><em> of size </em><code>queries.length</code><em>, where </em><code>ans[i]</code><em> is the answer to the </em><code>i</code><sup>th</sup><em> query</em>.</p>
+
+<p><strong>Note</strong> that the vowel letters are <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>.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;aba&quot;,&quot;bcb&quot;,&quot;ece&quot;,&quot;aa&quot;,&quot;e&quot;], queries = [[0,2],[1,4],[1,1]]
+<strong>Output:</strong> [2,3,0]
+<strong>Explanation:</strong> The strings starting and ending with a vowel are &quot;aba&quot;, &quot;ece&quot;, &quot;aa&quot; and &quot;e&quot;.
+The answer to the query [0,2] is 2 (strings &quot;aba&quot; and &quot;ece&quot;).
+to query [1,4] is 3 (strings &quot;ece&quot;, &quot;aa&quot;, &quot;e&quot;).
+to query [1,1] is 0.
+We return [2,3,0].
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+
+<pre>
+<strong>Input:</strong> words = [&quot;a&quot;,&quot;e&quot;,&quot;i&quot;], queries = [[0,2],[0,1],[2,2]]
+<strong>Output:</strong> [3,2,1]
+<strong>Explanation:</strong> Every string satisfies the conditions, so we return [3,2,1].</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li><code>1 &lt;= words.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= words[i].length &lt;= 40</code></li>
+	<li><code>words[i]</code> consists only of lowercase English letters.</li>
+	<li><code>sum(words[i].length) &lt;= 3 * 10<sup>5</sup></code></li>
+	<li><code>1 &lt;= queries.length &lt;= 10<sup>5</sup></code></li>
+	<li><code>0 &lt;= l<sub>i</sub> &lt;= r<sub>i</sub> &lt;&nbsp;words.length</code></li>
+</ul>

2691-count-vowel-strings-in-ranges/solution.py

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+# Approach: Prefix Sum
+
+# M = size of `words`, N = size of `queries`
+# Time: O(M + N)
+# Space: O(M)
+
+class Solution:
+    def vowelStrings(self, words: List[str], queries: List[List[int]]) -> List[int]:
+        ans = [0] * len(queries)
+        vowels = ['a', 'e', 'i', 'o', 'u']
+        prefix_sum = [0] * len(words)
+        sum = 0
+
+        for i in range(len(words)):
+            current_word = words[i]
+            if (current_word[0] in vowels and current_word[-1] in vowels):
+                sum += 1
+            prefix_sum[i] = sum
+
+        for i in range(len(queries)):
+            current_query = queries[i]
+            ans[i] = prefix_sum[current_query[1]] - (0 if current_query[0] == 0 else prefix_sum[current_query[0] - 1])
+
+        return ans
+        
+
+