Create: 1456 - Maximum Number of Vowels in a Substring of Given Length

javascript/1456-maximum-number-of-vowels-in-a-substring-of-given-length.js

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+/**
+ * @param {string} s
+ * @param {number} k
+ * @return {number}
+ */
+
+// Time complexity: O(N)
+// Space complexity: O(1)
+
+var maxVowels = function (s, k) {
+    const vowels = new Set(['a', 'e', 'i', 'o', 'u']);
+    let left = 0;
+    let count = 0;
+    let result = 0;
+
+    for (let right = 0; right < s.length; right++) {
+        count += vowels.has(s[right]) ? 1 : 0;
+
+        if (right - left + 1 > k) {
+            count -= vowels.has(s[left]) ? 1 : 0;
+            left++;
+        }
+
+        result = Math.max(result, count);
+    }
+
+    return result;
+};

python/1456-maximum-number-of-vowels-in-a-substring-of-given-length.py

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+# Time complexity: O(N)
+# Space complexity: O(1)
+
+
+class Solution:
+    def maxVowels(self, s: str, k: int) -> int:
+        vowels = {"a", "e", "i", "o", "u"}
+
+        left, count, result = 0, 0, 0
+
+        for right in range(len(s)):
+            count += 1 if s[right] in vowels else 0
+
+            if right - left + 1 > k:
+                count -= 1 if s[left] in vowels else 0
+                left += 1
+
+            result = max(result, count)
+
+        return result

typescript/1456-maximum-number-of-vowels-in-a-substring-of-given-length.ts

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+// Time complexity: O(N)
+// Space complexity: O(1)
+
+function maxVowels(s: string, k: number): number {
+    const vowels: Set<string> = new Set(['a', 'e', 'i', 'o', 'u']);
+    let left: number = 0;
+    let count: number = 0;
+    let result: number = 0;
+
+    for (let right: number = 0; right < s.length; right++) {
+        count += vowels.has(s[right]) ? 1 : 0;
+
+        if (right - left + 1 > k) {
+            count -= vowels.has(s[left]) ? 1 : 0;
+            left++;
+        }
+
+        result = Math.max(result, count);
+    }
+
+    return result;
+}