2962. Count Subarrays Where Max Element Appears at Least K Times - Explanation

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        n, res = len(nums), 0
        maxi = max(nums)

        for i in range(n):
            cnt = 0
            for j in range(i, n):
                if nums[j] == maxi:
                    cnt += 1

                if cnt >= k:
                    res += 1

        return res

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        max_n, max_cnt = max(nums), 0
        l = 0
        res = 0

        for r in range(len(nums)):
            if nums[r] == max_n:
                max_cnt += 1

            while max_cnt > k or (l <= r and max_cnt == k and nums[l] != max_n):
                if nums[l] == max_n:
                    max_cnt -= 1
                l += 1

            if max_cnt == k:
                res += l + 1

        return res

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        max_n, max_cnt = max(nums), 0
        l = res = 0

        for r in range(len(nums)):
            if nums[r] == max_n:
                max_cnt += 1
            while max_cnt == k:
                if nums[l] == max_n:
                    max_cnt -= 1
                l += 1
            res += l

        return res

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        n = len(nums)
        max_n = max(nums)
        max_indexes = [-1]
        for i, num in enumerate(nums):
            if num == max_n:
                max_indexes.append(i)

        res = 0
        for i in range(1, len(max_indexes) - k + 1):
            cur = (max_indexes[i] - max_indexes[i - 1])
            cur *= (n - max_indexes[i + k - 1])
            res += cur

        return res

class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        max_n = max(nums)
        max_indexes = deque()
        res = 0

        for i, num in enumerate(nums):
            if num == max_n:
                max_indexes.append(i)

            if len(max_indexes) > k:
                max_indexes.popleft()

            if len(max_indexes) == k:
                res += max_indexes[0] + 1

        return res