leetcode1297

class Solution:
    def maxFreq(self, s: str, maxLetters: int, minSize: int, maxSize: int) -> int:
        n = len(s)
        if n < minSize:
            return 0
        left,right = 0,0
        #dic = {}
        res = {}
        while left <= n - minSize:
            for width in range(minSize,maxSize+1):
                right = left + width
                if right > n:
                    break
                #print('',width,'left:',left,'right:',right)
                dic = {}
                diffcount = 0
                for i in range(left,right):
                    cur = s[i]
                    if cur not in dic:
                        dic[cur] = 1
                        diffcount += 1
                    else:
                        dic[cur] += 1
                if diffcount <= maxLetters:
                    word = s[left:right]
                    if word not in res:
                        res[word] = 1
                    else:
                        res[word] += 1
                else:
                    break
            left += 1
        if len(res) > 0:
            return max(res.values())
        else:
            return 0

算法思路：滑动窗口。

窗口的范围是[minSize,maxSize]之间，left从左到右遍历，每次得到一个right。

统计left到right这个窗口内的元素，有多少个不同元素，是否满足maxLetters限制。

如果满足条件则存储进res中，最后返回res出现次数最多的值(value)。