LeetCode 2348. Number of Zero-Filled Subarrays Solution in Java, C++, Python & More | Explanation + Code

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

After the traversal, we return the answer.

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

Similar problems:

• 413. Arithmetic Slices
• 1513. Number of Substrings With Only 1s

PythonJavaC++GoTypeScriptRust

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(code-box)

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; } }(code-box)

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; } };(code-box)

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

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

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 } }(code-box)