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LiveCodeBench / 2866 / longest-even-odd-subarray-with-threshold
Problem
Scored by the benchmark’s own harness. Use the benchmark’s own evaluation harness to check your work.
contest date
2023-07-02T00:00:00
contest id
weekly-contest-352
difficulty
easy
platform
leetcode
question content
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
question title
longest-even-odd-subarray-with-threshold
starter code
Code
class Solution:
def longestAlternatingSubarray(self, nums: List[int], threshold: int) -> int:
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initial import