An array is considered special if the parity of every pair of adjacent elements is different. In other words, one element in each pair must be even, and the other must be odd.
You are given an array of integers nums. Return true if nums is a special array, otherwise, return false.
Example 1:
Input: nums = [1]
Output: true
Explanation:
There is only one element. So the answer is true.
Example 2:
Input: nums = [2,1,4]
Output: true
Explanation:
There is only two pairs: (2,1) and (1,4), and both of them contain numbers with different parity. So the answer is true.
Example 3:
Input: nums = [4,3,1,6]
Output: false
Explanation:
nums[1] and nums[2] are both odd. So the answer is false.
Constraints:
1 <= nums.length <= 1001 <= nums[i] <= 100When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:
The brute force approach involves guessing a special number and checking if it meets the problem's condition. We will try every possible number in a range until we find the special number or exhaust all options.
Here's how the algorithm would work step-by-step:
def special_array_i_brute_force(numbers):
array_length = len(numbers)
for potential_special_value in range(array_length + 1):
# Count elements greater or equal to the potential value
count = 0
for number in numbers:
if number >= potential_special_value:
count += 1
# Check if potential value is the special number
if count == potential_special_value:
return potential_special_value
# No special value found, after checking
# all potential values.
return -1The problem asks us to find a special number within a list. A number is 'special' if exactly that many numbers in the list are greater than or equal to it. The optimal approach efficiently searches for this special number without needing to check every possibility by utilizing a clever strategy of elimination.
Here's how the algorithm would work step-by-step:
def find_special_integer(numbers):
numbers.sort()
list_length = len(numbers)
left_index = 0
right_index = list_length - 1
while left_index <= right_index:
potential_special_number = (left_index + right_index) // 2
count = 0
# Count elements greater or equal
for number in numbers:
if number >= numbers[potential_special_number]:
count += 1
# Check if it's the special integer
if count == numbers[potential_special_number]:
return numbers[potential_special_number]
# Adjust the search range if it is too low
if count < numbers[potential_special_number]:
right_index = potential_special_number - 1
# Adjust the search range if it is too high
else:
left_index = potential_special_number + 1
return -1| Case | How to Handle |
|---|---|
| Null or empty input array | Return 0 immediately as there's no array to process. |
| Input array with a single element | Return 0 immediately as no 'special' number can exist. |
| Array with all elements being 0 | The algorithm should iterate and count elements correctly, correctly returning the 'special' number or 0. |
| Array sorted in descending order | The algorithm should handle this without issues by iterating and comparing elements with their index. |
| Array with very large integers | The algorithm compares the integer values with their index so these should not affect complexity or cause issues if within acceptable bounds for the programming language. |
| Array containing duplicates and a 'special' number exists | The 'special' number should still be found correctly despite duplicates, if it exists. |
| No 'special' number exists in the input | The algorithm should correctly return 0 after checking all possibilities. |
| Array of maximum size allowed by memory constraints | The algorithm must iterate through the array in O(n) time and not allocate unnecessary memory to avoid memory constraints. |