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LeetCode 2760: Longest Even Odd Subarray With Threshold Solution

Master LeetCode problem 2760 (Longest Even Odd Subarray With Threshold), a easy challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

2760. Longest Even Odd Subarray With Threshold

Easy
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Problem Explanation

Explanation:

To solve this problem, we can iterate through the array while keeping track of the current subarray that satisfies the given conditions. We will update the subarray length whenever we find a longer subarray that meets the conditions. We can use two pointers to keep track of the current subarray.

Algorithm:

  1. Initialize variables start and end to 0, representing the start and end indices of the current subarray, and maxLength to 0, representing the length of the longest subarray found.
  2. Iterate over the array using a loop:
    • If the current element is odd or greater than the threshold, update start to the index of the next element.
    • If the current element is even and less than or equal to the threshold, update end to the current index.
    • Calculate the length of the current subarray as end - start + 1 and update maxLength if it is greater than the current maxLength.
  3. Return maxLength.

Time Complexity: O(N) where N is the number of elements in the array. Space Complexity: O(1)

Solution Code

class Solution {
    public int longestSubarray(int[] nums, int threshold) {
        int maxLength = 0;
        int start = 0, end = 0;
        
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] % 2 != 0 || nums[i] > threshold) {
                start = i + 1;
            } else {
                end = i;
                maxLength = Math.max(maxLength, end - start + 1);
            }
        }
        
        return maxLength;
    }
}

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