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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
public long zeroFilledSubarray(int[] nums) {
long res = 0;
for (int i = 0, j = 0; i < nums.length; ++i) {
if (nums[i] != 0)
j = i + 1;
res += i - j + 1;
}
return res;
}
long long zeroFilledSubarray(vector<int>& nums) {
long long res = 0;
for (int i = 0, j = 0; i < nums.size(); ++i) {
if (nums[i] != 0)
j = i + 1;
res += i - j + 1;
}
return res;
}
class Solution:
def zeroFilledSubarray(self, nums: List[int]) -> int:
n = len(nums)
ans = 0
i, j = 0, 0
while i <= n - 1:
j = 0
if nums[i] == 0:
while i + j <= n - 1 and nums[i + j] == 0:
j += 1
ans += (j + 1) * j // 2
i = i + j + 1
return ans
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