How do you solve Arithmetic Subarrays on LeetCode?

Arithmetic Subarrays (LeetCode #1630) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Sorting the subarray is essential to easily check for arithmetic properties.

What is the time complexity of Arithmetic Subarrays?

The optimal solution for Arithmetic Subarrays runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to sorting the subarray, while the space complexity remains O(n) for storing the subarray.

What is the brute force solution for Arithmetic Subarrays?

The brute force approach for Arithmetic Subarrays is Brute Force, with O(n² log n) time complexity and O(n) space complexity. The brute-force approach involves checking every possible subarray defined by the given ranges and determining if it can be rearranged into an arithmetic sequence. This is done by sorting the subarray and checking the differences between consecutive elements. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Arithmetic Subarrays?

Arithmetic Subarrays uses the following concepts: Array, Hash Table, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Arithmetic Subarrays?

Always clarify the problem constraints and edge cases with the interviewer. Think aloud while solving to demonstrate your thought process. Consider both time and space complexity when discussing your solution.

What are common mistakes when solving Arithmetic Subarrays?

Not sorting the subarray before checking differences. Failing to handle edge cases, such as subarrays of size 2.

#1630

Arithmetic Subarrays

Medium

ArrayHash TableSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

The optimal solution leverages the fact that a sorted sequence can be checked for arithmetic properties more efficiently. By sorting the subarray and checking the differences in one pass, we reduce the overall complexity significantly.

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Algorithm

5 steps

1Step 1: For each query, extract the subarray defined by the indices l[i] and r[i].
2Step 2: Sort the extracted subarray.
3Step 3: Calculate the common difference using the first two elements.
4Step 4: Use a single loop to check if all consecutive differences are equal to the common difference.
5Step 5: Append the result to the answer list.

solution.py8 lines

1def checkArithmeticSubarrays(nums, l, r):
2    answer = []
3    for start, end in zip(l, r):
4        subarray = sorted(nums[start:end + 1])
5        diff = subarray[1] - subarray[0]
6        is_arithmetic = all(subarray[i + 1] - subarray[i] == diff for i in range(1, len(subarray) - 1))
7        answer.append(is_arithmetic)
8    return answer

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Complexity note: The time complexity is O(n log n) due to sorting the subarray, while the space complexity remains O(n) for storing the subarray.

1Sorting the subarray is essential to easily check for arithmetic properties.
2The common difference must remain consistent across the entire sorted subarray.

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