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2348. Number of Zero-Filled Subarrays

Description

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

Solutions

Solution 1: Traversal and Counting

We traverse the array \(\textit{nums}\) and use a variable \(\textit{cnt}\) to record the current number of consecutive \(0\)s. For the current element \(x\) we are traversing, if \(x\) is \(0\), then \(\textit{cnt}\) is incremented by \(1\), and the number of all-zero subarrays ending with the current \(x\) is \(\textit{cnt}\), which we add to the answer. Otherwise, we set \(\textit{cnt}\) to \(0\).

After the traversal, we return the answer.

Time complexity \(O(n)\), where \(n\) is the length of the array \(\textit{nums}\). Space complexity \(O(1)\).

Similar problems:

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

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class Solution {
    public long zeroFilledSubarray(int[] nums) {
        long ans = 0;
        int cnt = 0;
        for (int x : nums) {
            if (x == 0) {
                ans += ++cnt;
            } else {
                cnt = 0;
            }
        }
        return ans;
    }
}

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class Solution {
public:
    long long zeroFilledSubarray(vector<int>& nums) {
        long long ans = 0;
        int cnt = 0;
        for (int x : nums) {
            if (x == 0) {
                ans += ++cnt;
            } else {
                cnt = 0;
            }
        }
        return ans;
    }
};

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

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function zeroFilledSubarray(nums: number[]): number {
    let [ans, cnt] = [0, 0];
    for (const x of nums) {
        if (!x) {
            ans += ++cnt;
        } else {
            cnt = 0;
        }
    }
    return ans;
}

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impl Solution {
    pub fn zero_filled_subarray(nums: Vec<i32>) -> i64 {
        let mut ans: i64 = 0;
        let mut cnt: i64 = 0;
        for x in nums {
            if x == 0 {
                cnt += 1;
                ans += cnt;
            } else {
                cnt = 0;
            }
        }
        ans
    }
}

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