2348. Number of Zero-Filled Subarrays #

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.```

<p><strong class="example">Example 2:</strong></p>

<p><strong class="example">Example 3:</strong></p>

<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
<li><code>-10<sup>9</sup> &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
</ul>


::: details _Click to open Hints_

- For each zero, you can calculate the number of zero-filled subarrays that end on that index, which is the number of consecutive zeros behind the current element + 1.
- Maintain the number of consecutive zeros behind the current element, count the number of zero-filled subarrays that end on each index, sum it up to get the answer.

:::

## Solution:

::: code-group

```rs [Rust]
impl Solution {
    pub fn zero_filled_subarray(nums: Vec<i32>) -> i64 {
        let mut nums = nums;
        nums.push(1);

        let (mut res, mut count) = (0, 0);
        for num in nums {
            if num == 0 {
                count += 1;
            } else {
                res += count * (count + 1) / 2;
                count = 0;
            }
        }

        res
    }
}