DDSA Solutions

3151. Special Array I

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Array
Time: O(n)
Space: O(1)

Problem Overview

Check if array is "special": every adjacent pair has different parity (one odd, one even).

Intuition

Check if array is "special": every adjacent pair has different parity (one odd, one even).

Algorithm

  1. 1For each i from 0 to n-2: if nums[i]%2 == nums[i+1]%2: return false.

Common Pitfalls

  • •Adjacent elements must alternate parity. One same-parity pair fails the check.
3151.cs
C#
// Approach: Check every adjacent pair has different parity.
// Time: O(n) Space: O(1)

public class Solution
{
    public bool IsArraySpecial(int[] nums)
    {
        for (int i = 1; i < nums.Length; ++i)
        {
            if (nums[i] % 2 == nums[i - 1] % 2)
                return false;
        }
        return true;
    }
}
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