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Leetcode 2760. Longest Even Odd Subarray With Threshold Sure, let’s tackle the LeetCode problem step-by-step.

Problem Statement

Given an array of integers nums sorted in non-decreasing order and an integer threshold, return the length of the longest subarray where:

  1. The subarray’s elements alternate between even and odd.
  2. All elements of the subarray are less than or equal to threshold.

Clarifying Questions

  1. Is the array always non-empty?
    • From the problem statement, let’s assume that the input array nums is non-empty.
  2. What should be returned if there’s no such subarray?
    • If no valid subarray exists, we should return 0.
  3. Can we have negative numbers in the array?
    • The problem statement doesn’t explicitly restrict that, so we will assume that nums can contain negative integers as well.
  4. What are the constraints on the size of the array?
    • Typically, LeetCode problems will have specific constraints on the size of the input (n), but we will assume n can be large, up to 10^5 or so, common in LeetCode problems.

Strategy

  1. Initialization:
    • Set up variables to keep track of the current subarray length (current_length) and the maximum subarray length (max_length).
  2. Iterate through the array:
    • Traverse through the array while checking for alternating even and odd elements.
    • Ensure each element is less than or equal to the threshold.
    • If the current element continues the even-odd pattern and is within the threshold, increment the current length.
    • If it breaks the pattern or exceeds the threshold, reset current_length.
  3. Update maximum length:
    • After each valid extension of the subarray, update max_length if current_length is greater than the current max_length.

Code Implementation

#include <vector>
#include <algorithm>

class Solution {
public:
    int longestEvenOddSubarray(std::vector<int>& nums, int threshold) {
        int n = nums.size();
        int max_length = 0;
        int current_length = 0;

        for (int i = 0; i < n; ++i) {
            if (nums[i] <= threshold) {
                if (current_length == 0) {
                    current_length = 1;
                } else if ((nums[i - 1] % 2 != nums[i] % 2)) {
                    current_length++;
                } else {
                    current_length = 1;
                }
                max_length = std::max(max_length, current_length);
            } else {
                current_length = 0;
            }
        }
        return max_length;
    }
};

Time Complexity

This solution efficiently finds the longest subarray with the described properties. The if conditions handle transitions between even and odd numbers and check for the threshold constraint appropriately.

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