How do you solve Number of Zero-Filled Subarrays on LeetCode?

Number of Zero-Filled Subarrays (LeetCode #2348) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Counting consecutive zeros allows us to calculate the number of zero-filled subarrays efficiently.

What is the time complexity of Number of Zero-Filled Subarrays?

The optimal solution for Number of Zero-Filled Subarrays runs in O(n) time and uses O(1) space. This is linear time complexity because we only traverse the array once, making it much more efficient than the brute-force approach.

What is the brute force solution for Number of Zero-Filled Subarrays?

The brute force approach for Number of Zero-Filled Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible subarray to see if it's filled with zeros. This is straightforward but inefficient, especially for large arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Zero-Filled Subarrays?

Number of Zero-Filled Subarrays uses the following concepts: Array, Math. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Number of Zero-Filled Subarrays?

Always consider edge cases, such as arrays with no zeros or all zeros. Explain your thought process clearly while coding, as it helps interviewers understand your approach. Practice identifying patterns in problems, as many can be solved with similar techniques.

What are common mistakes when solving Number of Zero-Filled Subarrays?

Not resetting the consecutive zero counter when encountering a non-zero element. Overcomplicating the problem by trying to generate all subarrays instead of counting.

#2348

Number of Zero-Filled Subarrays

Medium

ArrayMathArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

Instead of checking every subarray, we can count consecutive zeros and calculate the number of zero-filled subarrays that can be formed from them. This is efficient because it reduces the problem to a single pass through the array.

⚙️

Algorithm

4 steps

1Step 1: Initialize a count variable to zero and a consecutive zero counter to zero.
2Step 2: Traverse the array. For each zero encountered, increment the consecutive zero counter.
3Step 3: When a non-zero is encountered, add the count of subarrays formed by the consecutive zeros to the total count.
4Step 4: Reset the consecutive zero counter when a non-zero is found.

solution.py10 lines

1def zeroFilledSubarray(nums):
2    count = 0
3    consecutive_zeros = 0
4    for num in nums:
5        if num == 0:
6            consecutive_zeros += 1
7            count += consecutive_zeros
8        else:
9            consecutive_zeros = 0
10    return count

ℹ

Complexity note: This is linear time complexity because we only traverse the array once, making it much more efficient than the brute-force approach.

1Counting consecutive zeros allows us to calculate the number of zero-filled subarrays efficiently.
2Each group of consecutive zeros contributes to multiple subarrays, which can be summed up.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.