package leetcode.p1456;

import java.util.HashSet;
import java.util.Set;

/**
 * @ProjectName: LeetCode
 * @FileName: Solution
 * @Author: 杨逸
 * @Data:2026/3/8 15:35
 * @Description: https://leetcode.cn/problems/maximum-number-of-vowels-in-a-substring-of-given-length/description/
 * 1456. 定长子串中元音的最大数目
 */
public class Solution {
    public int maxVowels(String s, int k) {
        //定长滑动窗口解法
        //关注进入窗口的字符和离开窗口的字符
        Set<Character> set = new HashSet<>();
        set.add('a');
        set.add('e');
        set.add('i');
        set.add('o');
        set.add('u');
        int count = 0;
        int ans = 0;
        for (int i = 0; i < s.length(); i++) {
            if (i >= k) {
                //离开窗口的字符
                if (set.contains(s.charAt(i - k))) count--;
            }
            //进入窗口的字符
            if (set.contains(s.charAt(i))) count++;
            ans = Math.max(ans, count);
        }
        return ans;
    }

    /**
     * 前缀和,计算最大期间差
     * 前缀和统计当前长度已经出现多少个元音字符
     * 以k为长度的滑动窗口，统计窗口内元音字符的个数
     * @param s
     * @param k
     * @return
     */
    private int prefixSolution(String s, int k){
        //前缀和
        int[] prefix = new int[s.length()+1];
        int count = 0;
        int max = 0;
        Set<Character> set = new HashSet<>();
        set.add('a');
        set.add('e');
        set.add('i');
        set.add('o');
        set.add('u');
        for (int i = 0; i < s.length(); i++) {
            if (set.contains(s.charAt(i)))count++;
            prefix[i+1] = count;
        }
        for (int i = k; i < prefix.length; i++) {
            if (prefix[i]- prefix[i-k] > max)max = prefix[i] - prefix[i-k];
        }
        return max;
    }
}