lc 395. Longest Substring with At Least K Repeating Characters

Find the length of the longest substring T of a given string (consists of lowercase letters only) such that every character in T appears no less than k times.

Example 1:

Input:
s = "aaabb", k = 3

Output:
3

The longest substring is "aaa", as 'a' is repeated 3 times.

Example 2:

Input:
s = "ababbc", k = 2

Output:
5

The longest substring is "ababb", as 'a' is repeated 2 times and 'b' is repeated 3 times.



　　题目要求最长，若不是因为部分字母不足k个，我们就输出整个字符串长度作为答案了。为什么不能呢，因为部分字母压根不足k个，假设它是字母‘a’。
那么不管我们的目标字符串长什么样，它必然不包含‘a’。那么我们可以用字母a作为分割符，把字符串分成若干段，继续进行判断。因为每次都是找到一些使当前判断字符串不成立的字母，
这样字母最多26个，所以26层循环就可以了。我门不断细分这些字符串，直到其长度为0或者满足条件记录长度。
　　判断满足性时间常数的，用一个大二维数组记录每个字母在每个位置出现次数就ok。

代码：

class Solution:
    def __init__(self):
        self.d=None
        self.k=0
    def getNum(self,l,r):
        if l==0:
            return self.d[r][:]
        ans=[0]*26
        for i in range(26):
            ans[i]=self.d[r][i]-self.d[l-1][i]
        return ans
    def noValid(self,a):
        ans=set()
        for i in range(26):
            if a[i]>0 and a[i]<self.k:
                ans.add(i)
        return ans
    def longestSubstring(self, s, k):
        """
        :type s: str
        :type k: int
        :rtype: int
        """
        self.k=k
        l=len(s)
        if l==0:
            return 0
        d=[[0]*26]
        d[-1][ord(s[0])-ord('a')]+=1
        for i in range(1,l):
            d.append(d[-1][:])
            d[-1][ord(s[i])-ord('a')]+=1
        self.d=d

        next=[(0,l-1)]
        maxx=0
        for _ in range(26):
            tem=[]
            for x,y in next:
                if x>y:
                    continue
                n=self.getNum(x,y)
                nov=self.noValid(n)
                if len(nov)==0:
                    maxx=max(maxx,y-x+1)
                    continue
                last=x
                for i in range(x,y+1):
                    if ord(s[i])-ord('a') in nov:
                        tem.append((last,i-1))
                        last=i+1
                tem.append((last,y))
            next=tem
        return maxx