README_EN.md

2760. Longest Even Odd Subarray With Threshold

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Description

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

Solutions

Python3

class Solution:
    def longestAlternatingSubarray(self, nums: List[int], threshold: int) -> int:
        ans, n = 0, len(nums)
        for l in range(n):
            if nums[l] % 2 == 0 and nums[l] <= threshold:
                r = l + 1
                while r < n and nums[r] % 2 != nums[r - 1] % 2 and nums[r] <= threshold:
                    r += 1
                ans = max(ans, r - l)
        return ans

Java

class Solution {
    public int longestAlternatingSubarray(int[] nums, int threshold) {
        int ans = 0, n = nums.length;
        for (int l = 0; l < n; ++l) {
            if (nums[l] % 2 == 0 && nums[l] <= threshold) {
                int r = l + 1;
                while (r < n && nums[r] % 2 != nums[r - 1] % 2 && nums[r] <= threshold) {
                    ++r;
                }
                ans = Math.max(ans, r - l);
            }
        }
        return ans;
    }
}

C++

class Solution {
public:
    int longestAlternatingSubarray(vector<int>& nums, int threshold) {
        int ans = 0, n = nums.size();
        for (int l = 0; l < n; ++l) {
            if (nums[l] % 2 == 0 && nums[l] <= threshold) {
                int r = l + 1;
                while (r < n && nums[r] % 2 != nums[r - 1] % 2 && nums[r] <= threshold) {
                    ++r;
                }
                ans = max(ans, r - l);
            }
        }
        return ans;
    }
};

Go

func longestAlternatingSubarray(nums []int, threshold int) int {
	ans, n := 0, len(nums)
	for l := range nums {
		if nums[l]%2 == 0 && nums[l] <= threshold {
			r := l + 1
			for r < n && nums[r]%2 != nums[r-1]%2 && nums[r] <= threshold {
				r++
			}
			ans = max(ans, r-l)
		}
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

TypeScript

function longestAlternatingSubarray(nums: number[], threshold: number): number {
    const n = nums.length;
    let ans = 0;
    for (let l = 0; l < n; ++l) {
        if (nums[l] % 2 === 0 && nums[l] <= threshold) {
            let r = l + 1;
            while (r < n && nums[r] % 2 !== nums[r - 1] % 2 && nums[r] <= threshold) {
                ++r;
            }
            ans = Math.max(ans, r - l);
        }
    }
    return ans;
}

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