How do you solve Special Array I on LeetCode?

Special Array I (LeetCode #3151) 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: Adjacent elements must have different parities.

What is the brute force solution for Special Array I?

The brute force approach for Special Array I is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach checks each pair of adjacent elements to see if they have different parities. If any adjacent elements share the same parity, the array is not special. The optimal Optimal Solution approach improves this to O(n).

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

Special Array I uses the following concepts: Array. The recommended patterns to study are: Array.

What are the interview tips for Special Array I?

Always clarify edge cases like single-element arrays. Explain your thought process while coding. Consider both time and space complexity in your solution.

What are common mistakes when solving Special Array I?

Not considering single-element arrays as special. Confusing parity checks (even vs odd).

#3151

Special Array I

Easy

ArrayArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal solution is similar to the brute force approach but emphasizes the fact that we can check for parity in a single pass without needing to store any additional information.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable to track the expected parity (even or odd) based on the first element.
2Step 2: Iterate through the array starting from the second element.
3Step 3: For each element, check if its parity matches the expected parity. If it does, return false.
4Step 4: Update the expected parity for the next iteration.
5Step 5: If the loop completes without returning false, return true.

solution.py9 lines

1# Full working Python code
2
3def is_special_array(nums):
4    expected_parity = nums[0] % 2
5    for i in range(1, len(nums)):
6        if (nums[i] % 2) == expected_parity:
7            return False
8        expected_parity = 1 - expected_parity  # Toggle parity
9    return True

ℹ

Complexity note: This complexity is O(n) because we still make a single pass through the array. Space complexity is O(1) since no additional data structures are used.

1Adjacent elements must have different parities.
2The parity can be toggled to simplify checks.

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