LeetCode-in-Java
2760. Longest Even Odd Subarray With Threshold
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
iin the range[l, r - 1],nums[i] % 2 != nums[i + 1] % 2 - For all indices
iin 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 <= 1001 <= nums[i] <= 1001 <= threshold <= 100
Solution
public class Solution {
public int longestAlternatingSubarray(int[] nums, int threshold) {
int maxLength = 0;
int i = 0;
while (i < nums.length) {
if (nums[i] % 2 == 0 && nums[i] <= threshold) {
int length = 1;
int j = i + 1;
while (j < nums.length && nums[j] <= threshold && nums[j] % 2 != nums[j - 1] % 2) {
length++;
j++;
}
maxLength = Math.max(maxLength, length);
i = j - 1;
}
i++;
}
return maxLength;
}
}