Class Solution
Easy
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
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Constructor Summary
Constructors -
Method Summary
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Constructor Details
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Solution
public Solution()
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Method Details
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longestAlternatingSubarray
public int longestAlternatingSubarray(int[] nums, int threshold)
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