2348. Number of Zero-Filled Subarrays - Explanation

1. Brute Force

class Solution:
    def zeroFilledSubarray(self, nums: List[int]) -> int:
        res = 0
        for i in range(len(nums)):
            for j in range(i, len(nums)):
                if nums[j] != 0:
                    break
                res += 1
        return res

Time & Space Complexity

Time complexity: $O(n ^ 2)$
Space complexity: $O(1)$

2. Count Consecutive Zeros - I

class Solution:
    def zeroFilledSubarray(self, nums: List[int]) -> int:
        res = i = 0
        while i < len(nums):
            count = 0
            while i < len(nums) and nums[i] == 0:
                count += 1
                i += 1
                res += count
            i += 1
        return res

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

3. Count Consecutive Zeros - II

class Solution:
    def zeroFilledSubarray(self, nums: List[int]) -> int:
        res = count = 0

        for num in nums:
            if num == 0:
                count += 1
            else:
                count = 0
            res += count

        return res

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$

4. Count Consecutive Zeros (Math)

class Solution:
    def zeroFilledSubarray(self, nums: List[int]) -> int:
        res = count = 0
        for num in nums:
            if num == 0:
                count += 1
            else:
                res += (count * (count + 1)) // 2
                count = 0
        res += (count * (count + 1)) // 2
        return res

Time & Space Complexity

Time complexity: $O(n)$
Space complexity: $O(1)$