Leetcode_2760_Longest_Even_Odd_Subarray_With_Threshold.py

"""
You are given a 0-indexed integer array nums and an integer threshold.

Find the length of the longest subarray of nums starting at index l and ending at index r (0 <= l <= r < nums.length) that satisfies the following conditions:

nums[l] % 2 == 0
For all indices i in the range [l, r - 1], nums[i] % 2 != nums[i + 1] % 2
For all indices i in the range [l, r], nums[i] <= threshold
Return an integer denoting the length of the longest such subarray.

Note: A subarray is a contiguous non-empty sequence of elements within an array.

 

Example 1:

Input: nums = [3,2,5,4], threshold = 5
Output: 3
Explanation: In this example, we can select the subarray that starts at l = 1 and ends at r = 3 => [2,5,4]. This subarray satisfies the conditions.
Hence, the answer is the length of the subarray, 3. We can show that 3 is the maximum possible achievable length.
Example 2:

Input: nums = [1,2], threshold = 2
Output: 1
Explanation: In this example, we can select the subarray that starts at l = 1 and ends at r = 1 => [2]. 
It satisfies all the conditions and we can show that 1 is the maximum possible achievable length.
Example 3:

Input: nums = [2,3,4,5], threshold = 4
Output: 3
Explanation: In this example, we can select the subarray that starts at l = 0 and ends at r = 2 => [2,3,4]. 
It satisfies all the conditions.
Hence, the answer is the length of the subarray, 3. We can show that 3 is the maximum possible achievable length.
 

Constraints:

1 <= nums.length <= 100 
1 <= nums[i] <= 100 
1 <= threshold <= 100

"""

#MY SOLUTION

class Solution:
    def longestAlternatingSubarray(self, nums: List[int], threshold: int) -> int:
        i = 0
        sz = maxi = 0
        
        while i<len(nums):
            if nums[i]%2==0 and nums[i]<=threshold:
                start = i
                while i < len(nums)-1 and nums[i] % 2 != nums[i + 1] % 2 and nums[i+1]<=threshold:
                    i += 1
                sz = i-start+1
                maxi = max(maxi,sz)
            i += 1
        
        return maxi