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231 lines (191 loc) · 5.68 KB
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Medium
1263
Weekly Contest 190 Q2
String
Sliding Window

中文文档

Description

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 <= 105
  • s consists of lowercase English letters.
  • 1 <= k <= s.length

Solutions

Solution 1: Sliding Window

First, we count the number of vowels in the first k characters, denoted as c n t , and initialize the answer a n s as c n t .

Then we start traversing the string from k . For each iteration, we add the current character to the window. If the current character is a vowel, we increment c n t . We remove the first character from the window. If the removed character is a vowel, we decrement c n t . Then, we update the answer a n s = max ( a n s , c n t ) .

After the traversal, we return the answer.

The time complexity is O ( n ) , where n is the length of the string s . The space complexity is O ( 1 ) .

Python3

class Solution:
    def maxVowels(self, s: str, k: int) -> int:
        vowels = set("aeiou")
        ans = cnt = sum(c in vowels for c in s[:k])
        for i in range(k, len(s)):
            cnt += int(s[i] in vowels) - int(s[i - k] in vowels)
            ans = max(ans, cnt)
        return ans

Java

class Solution {
    public int maxVowels(String s, int k) {
        int cnt = 0;
        for (int i = 0; i < k; ++i) {
            if (isVowel(s.charAt(i))) {
                ++cnt;
            }
        }
        int ans = cnt;
        for (int i = k; i < s.length(); ++i) {
            if (isVowel(s.charAt(i))) {
                ++cnt;
            }
            if (isVowel(s.charAt(i - k))) {
                --cnt;
            }
            ans = Math.max(ans, cnt);
        }
        return ans;
    }

    private boolean isVowel(char c) {
        return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
    }
}

C++

class Solution {
public:
    int maxVowels(string s, int k) {
        auto isVowel = [](char c) {
            return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
        };
        int cnt = count_if(s.begin(), s.begin() + k, isVowel);
        int ans = cnt;
        for (int i = k; i < s.size(); ++i) {
            cnt += isVowel(s[i]) - isVowel(s[i - k]);
            ans = max(ans, cnt);
        }
        return ans;
    }
};

Go

func maxVowels(s string, k int) int {
	isVowel := func(c byte) bool {
		return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u'
	}
	cnt := 0
	for i := 0; i < k; i++ {
		if isVowel(s[i]) {
			cnt++
		}
	}
	ans := cnt
	for i := k; i < len(s); i++ {
		if isVowel(s[i-k]) {
			cnt--
		}
		if isVowel(s[i]) {
			cnt++
		}
		ans = max(ans, cnt)
	}
	return ans
}

TypeScript

function maxVowels(s: string, k: number): number {
    const vowels = new Set(['a', 'e', 'i', 'o', 'u']);
    let cnt = 0;
    for (let i = 0; i < k; i++) {
        if (vowels.has(s[i])) {
            cnt++;
        }
    }
    let ans = cnt;
    for (let i = k; i < s.length; i++) {
        if (vowels.has(s[i])) {
            cnt++;
        }
        if (vowels.has(s[i - k])) {
            cnt--;
        }
        ans = Math.max(ans, cnt);
    }
    return ans;
}

PHP

class Solution {
    /**
     * @param String $s
     * @param Integer $k
     * @return Integer
     */
    function isVowel($c) {
        return $c === 'a' || $c === 'e' || $c === 'i' || $c === 'o' || $c === 'u';
    }
    function maxVowels($s, $k) {
        $cnt = 0;
        for ($i = 0; $i < $k; $i++) {
            if ($this->isVowel($s[$i])) {
                $cnt++;
            }
        }
        $ans = $cnt;
        for ($j = $k; $j < strlen($s); $j++) {
            if ($this->isVowel($s[$j - $k])) {
                $cnt--;
            }
            if ($this->isVowel($s[$j])) {
                $cnt++;
            }
            $ans = max($ans, $cnt);
        }
        return $ans;
    }
}