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, 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

    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