6293. Count the number of good array

2024-01-01

字数统计: 187字 | 阅读时长: 1min

At first glance, this question is a double pointer，But I am not familiar with the double pointer，Most of the time was wasted when you set up the framework。

This question is the third question of Zhou Sai，The first two questions were killed quickly，But because I slept，It’s only half an hour left when the computer is turned on。

always theredebug，Watch this afternoon0x3fSolution。There is a set of explanations of his sliding window algorithm and double pointer。

from collections import Counter
from itertools import count
from typing import List


class Solution:
    def countGood(self, nums: List[int], k: int) -> int:
        start = 0
        tail = 0
        flag = 0
        if len(set(nums)) == 1:
            if k > len(nums) * (len(nums)-1) // 2:
                return 0
        while start < len(nums) - 1:
            x = Counter(nums[start:tail])
            if len(x) >= k:
                flag2 = 0
                for j in x:
                    if x[j] >= k:
                        flag2 += 1
                if flag2 >= k:
                    flag += 1
            else:
                for i in x:
                    if x[i] * (x[i] - 1) // 2 >= k:
                        flag += 1
            if tail != len(nums):
                tail += 1
            else:
                start += 1

        return flag