Longest Even Odd Subarray With Threshold

Problem

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

Solution (Java)

class Solution {
    public int longestAlternatingSubarray(int[] nums, int th) {
        int n = nums.length,maxi = 0,i = 0,j = 0,flag = 0;
        while(j < n){
            if(flag == 0){
               if(nums[j] % 2 == 0 && nums[j] <= th){
                    i = j;
                    maxi = Math.max(maxi,j - i + 1);
                    flag = 1;
                }
            }
            else if(flag == 1){
                int x = nums[j-1],y = nums[j],c = x + y;
                if(c % 2 != 0 && nums[j] <= th)
                    maxi = Math.max(maxi,j - i + 1);
                else{
                    flag = 0;
                    j--;
                }
            }
            j++;
        }
        return maxi;
    }
}

Explain:

nope.

Complexity:

Time complexity : O(n).
Space complexity : O(n).