How do you solve Special Array II on LeetCode?

Special Array II (LeetCode #3152) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(n) space complexity. Key insight: Identifying segments of special subarrays allows for efficient querying.

What is the brute force solution for Special Array II?

The brute force approach for Special Array II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each subarray defined by the queries to see if it meets the special condition. This means iterating through each query and checking every adjacent pair in the specified range. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Special Array II?

Special Array II uses the following concepts: Array, Binary Search, Prefix Sum. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Special Array II?

Always discuss your thought process and potential optimizations with the interviewer. Be mindful of edge cases and test your solution against them. Practice writing clean and efficient code, as readability matters in interviews.

What are common mistakes when solving Special Array II?

Not considering edge cases where the subarray has only one element. Failing to optimize the solution when the brute force method is too slow.

#3152

Special Array II

Medium

ArrayBinary SearchPrefix SumArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n + m)Space O(n)

Instead of checking each subarray for every query, we can preprocess the array to identify segments of special subarrays. This allows us to quickly determine if two indices belong to the same special segment.

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Algorithm

3 steps

1Step 1: Create an array to store the segment identifier for each index in nums.
2Step 2: Traverse nums and assign a segment number to each index based on whether it continues the special property from the previous index.
3Step 3: For each query, check if the segment identifiers of the two indices are the same.

solution.py20 lines

1def is_special_array(nums, queries):
2    n = len(nums)
3    segments = [0] * n
4    segment_id = 0
5    segments[0] = segment_id
6
7    for i in range(1, n):
8        if (nums[i] % 2) != (nums[i - 1] % 2):
9            segment_id += 1
10        segments[i] = segment_id
11
12    results = []
13    for from_i, to_i in queries:
14        results.append(segments[from_i] == segments[to_i])
15    return results
16
17# Example usage
18nums = [4, 3, 1, 6]
19queries = [[0, 2], [2, 3]]
20print(is_special_array(nums, queries))  # Output: [False, True]

ℹ

Complexity note: The time complexity is O(n + m) where n is the length of nums and m is the number of queries. We preprocess the array in O(n) and then answer each query in O(1).

1Identifying segments of special subarrays allows for efficient querying.
2Using parity checks can quickly determine if adjacent elements differ.

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