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Formatted question description: https://leetcode.ca/all/2348.html

# 2348. Number of Zero-Filled Subarrays

• Difficulty: Medium.
• Related Topics: Array, Math.
• Similar Questions: Arithmetic Slices, Number of Smooth Descent Periods of a Stock.

## Problem

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  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  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

## Solution (Java, C++, Python)

• class Solution {
public long zeroFilledSubarray(int[] nums) {
long cnt = 0L;
long local = 0L;
for (int n : nums) {
if (n == 0) {
cnt += ++local;
} else {
local = 0;
}
}
return cnt;
}
}

############

class Solution {
public long zeroFilledSubarray(int[] nums) {
long ans = 0;
int cnt = 0;
for (int v : nums) {
cnt = v != 0 ? 0 : cnt + 1;
ans += cnt;
}
return ans;
}
}

• class Solution:
def zeroFilledSubarray(self, nums: List[int]) -> int:
ans = cnt = 0
for v in nums:
cnt = 0 if v else cnt + 1
ans += cnt
return ans

############

# 2348. Number of Zero-Filled Subarrays
# https://leetcode.com/problems/number-of-zero-filled-subarrays/

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

for x in nums:
if x != 0:
curr = 0
else:
curr += 1
res += curr

return res


• class Solution {
public:
long long zeroFilledSubarray(vector<int>& nums) {
long long ans = 0;
int cnt = 0;
for (int& v : nums) {
cnt = v ? 0 : cnt + 1;
ans += cnt;
}
return ans;
}
};

• func zeroFilledSubarray(nums []int) (ans int64) {
cnt := 0
for _, v := range nums {
if v != 0 {
cnt = 0
} else {
cnt++
}
ans += int64(cnt)
}
return
}

• function zeroFilledSubarray(nums: number[]): number {
let ans = 0;
let cnt = 0;
for (const v of nums) {
cnt = v ? 0 : cnt + 1;
ans += cnt;
}
return ans;
}



Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).