根据大佬给出的O(n)做法的思路，提供一种python实现

def check(start, end):
    cnt = 0
    for i in range(26):
        if lst[i][end-1] - (lst[i][start-2] if start >= 2 else 0) == k:
            cnt += 1
    return cnt

n, k = map(int,input().split(" "))
inp = input()
lst = [[0] * n for i in range(26)]
#n列，26行(第一列为0)
for i in range(26):
    letter = chr(ord('a') + i)
    for j in range(len(inp)):
        if letter == inp[j]:
            lst[i][j] = lst[i][j-1] + 1
        else:
            lst[i][j] = lst[i][j-1]
ans = 0
for i in range(1,27):
    length = i * k
    for j in range(length,n + 1):
        l = j - length + 1
        r = j
        if check(l, r) == i:
            ans += 1
print(ans)