My first approach was to first check if the length of the array nums was 1. If it was, I will return True. Next, I set up a loop to iterate over every element in the array. Normally, I would expect to compare each pair of adjacent elements—comparing nums[i] with nums[i+1] to ensure that one is even and the other is odd. However, the code I wrote doesn’t actually compare adjacent pairs. Instead, within the loop, I use a condition that checks (nums[i] % 2 == 0 or nums[i] % 2 != 0) for each element. I realized that this condition is always true for any integer because every number is either even or odd. Since I repeat this check twice with an and between them, the overall condition remains always true.
class Solution:
def isArraySpecial(self, nums: List[int]) -> bool:
if len(nums) == 1:
return True
for i in range(len(nums)):
if (nums[i] % 2 == 0 or nums[i] % 2 != 0) and (nums[i] % 2 == 0 or nums[i] % 2 != 0):
return False
else:
return True- Time: O(n)
- Space: O(1)
My second approach was more inclusive. Instead of the unnecessary complex conditional, I iterate through the array from the first element up to the second-to-last element. For each index i, I compare the parity of nums[i] and nums[i + 1] by checking the remainder when divided by 2. If the remainder of both elements is the same, it means both numbers are either even or odd, which violates the requirement that adjacent elements have different parity. In that case, I return False. If I finish the loop without finding any adjacent pair with the same parity, I then return True.
class Solution:
def isArraySpecial(self, nums: List[int]) -> bool:
if len(nums) == 1:
return True
for i in range(len(nums) - 1):
if nums[i] % 2 == nums[i + 1] % 2:
return False
return True- Time: O(n)
- Space: O(1)
