Given a string s and an integer k, return the maximum number of vowel letters in any substring of s with length k.
Vowel letters in English are 'a', 'e', 'i', 'o', and 'u'.
Example 1:
Input: s = "abciiidef", k = 3 Output: 3 Explanation: The substring "iii" contains 3 vowel letters.
Example 2:
Input: s = "aeiou", k = 2 Output: 2 Explanation: Any substring of length 2 contains 2 vowels.
Example 3:
Input: s = "leetcode", k = 3 Output: 2 Explanation: "lee", "eet" and "ode" contain 2 vowels.
Constraints:
1 <= s.length <= 105s consists of lowercase English letters.1 <= k <= s.lengthThis approach uses the sliding window technique to efficiently find the maximum number of vowels in any substring of length k.
We first calculate the number of vowels in the initial window of length k. Then, for each subsequent character, we slide the window by one character to the right: we remove the character that goes out of the window and add the one that comes into the window, updating our vowel count accordingly.
This ensures that we examine each character in the string only once, leading to an O(n) time complexity.
The solution involves converting the character check into a vowel helper function to identify vowels. We initiate a loop through the first k characters to initialize our vowel count, then iterate over the remainder of the string, adjusting our count as we move the window along. The result is the maximum count found in any window. Utilizing helper functions helps improve code readability.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n), where n is the length of the string. We iterate through the string with a constant number of operations for each character.
Space Complexity: O(1), as we use a fixed amount of extra space.
Another way to tackle this problem is to use a combination of prefix sums and binary search to precompute regions of vowels.
First, we establish a prefix sum array that accumulates the number of vowels encountered at each character index in the string. Then, we use a binary search within this prefix sum to efficiently find the maximum number of vowels in any moving window of length k.
This approach is more complex and is generally less optimal than the sliding window approach but provides an alternative methodology.
The C solution creates a prefix sum array to accumulate the number of vowels found up to each point in the string. Using this prefix sum, the maximum number of vowels in any contiguous substring of length k can be determined by calculating the difference between pairs of prefix sums separated by k indices.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n), for prefix sum construction and analysis.
Space Complexity: O(n), for the prefix sum array.
| Approach | Complexity |
|---|---|
| Sliding Window Approach | Time Complexity: O(n), where n is the length of the string. We iterate through the string with a constant number of operations for each character. |
| Prefix Sum and Binary Search Approach | Time Complexity: O(n), for prefix sum construction and analysis. |
Maximum Number of Vowels in a Substring of Given Length - Leetcode 1456 - Python • NeetCodeIO • 17,445 views views
Watch 9 more video solutions →Practice Maximum Number of Vowels in a Substring of Given Length with our built-in code editor and test cases.
Practice on FleetCode