How do you solve Number of Centered Subarrays on LeetCode?

Number of Centered Subarrays (LeetCode #3804) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Single-element subarrays are always centered.

What is the time complexity of Number of Centered Subarrays?

The optimal solution for Number of Centered Subarrays runs in O(n) time and uses O(n) space. This complexity is due to a single pass through the array and the use of a hashmap for constant time lookups.

What is the brute force solution for Number of Centered Subarrays?

The brute force approach for Number of Centered Subarrays is Brute Force, with O(n²) time complexity and O(1) space complexity. Generate all possible subarrays and check if the sum of each subarray equals any of its elements. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Centered Subarrays?

Number of Centered Subarrays uses the following concepts: Array, Hash Table, Enumeration. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Number of Centered Subarrays?

Understand the problem requirements thoroughly. Practice identifying patterns in subarray problems. Discuss your thought process clearly during the interview.

What are common mistakes when solving Number of Centered Subarrays?

Not considering single-element subarrays. Incorrectly checking for sums in subarrays.

#3804

Number of Centered Subarrays

Medium

ArrayHash TableEnumerationHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Use a hashmap to track the sums of subarrays and check if any sum matches an element in the subarray efficiently.

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Algorithm

3 steps

1Step 1: Initialize a hashmap to store the frequency of sums and a variable to count centered subarrays.
2Step 2: Iterate through the array while maintaining a running sum and check if this sum exists in the hashmap.
3Step 3: For each element, check if the current sum matches any previous sums stored in the hashmap.

solution.py10 lines

1def count_centered_subarrays(nums):
2    count = 0
3    sum_map = {0: 1}
4    current_sum = 0
5    for num in nums:
6        current_sum += num
7        if current_sum in sum_map:
8            count += sum_map[current_sum]
9        sum_map[current_sum] = sum_map.get(current_sum, 0) + 1
10    return count

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Complexity note: This complexity is due to a single pass through the array and the use of a hashmap for constant time lookups.

1Single-element subarrays are always centered.
2The sum of a subarray can be efficiently tracked using a hashmap.

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