# Solving Leetcode Interviews in Seconds with AI: Number of Zero-Filled Subarrays


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2348" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given an integer array nums, return the number of subarrays filled with 0. A subarray is a contiguous non-empty sequence of elements within an array.   Example 1:  Input: nums = [1,3,0,0,2,0,0,4] Output: 6 Explanation:  There are 4 occurrences of [0] as a subarray. There are 2 occurrences of [0,0] as a subarray. There is no occurrence of a subarray with a size more than 2 filled with 0. Therefore, we return 6. Example 2:  Input: nums = [0,0,0,2,0,0] Output: 9 Explanation: There are 5 occurrences of [0] as a subarray. There are 3 occurrences of [0,0] as a subarray. There is 1 occurrence of [0,0,0] as a subarray. There is no occurrence of a subarray with a size more than 3 filled with 0. Therefore, we return 9.  Example 3:  Input: nums = [2,10,2019] Output: 0 Explanation: There is no subarray filled with 0. Therefore, we return 0.    Constraints:  1 <= nums.length <= 105 -109 <= nums[i] <= 109  

	# Explanation
	*   Iterate through the array, tracking the length of consecutive zeros.
*   When a non-zero element is encountered, calculate the number of subarrays formed by the preceding zeros and add it to the total.
*   Reset the zero-count and continue iterating.

*   Runtime Complexity: O(n), Storage Complexity: O(1)

	
	# Code
	```python
	def zero_filled_subarray(nums):
    """
    Given an integer array nums, return the number of subarrays filled with 0.

    Args:
        nums: An integer array.

    Returns:
        The number of subarrays filled with 0.
    """

    count = 0
    zero_length = 0
    for num in nums:
        if num == 0:
            zero_length += 1
        else:
            count += zero_length * (zero_length + 1) // 2
            zero_length = 0

    count += zero_length * (zero_length + 1) // 2  # Handle trailing zeros

    return count
	```