LeetCode #2760 — EASY

Longest Even Odd Subarray With Threshold

Build confidence with an intuition-first walkthrough focused on array fundamentals.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Sliding Window

Example 1

[3,2,5,4]
5

Example 2

[1,2]
2

Example 3

[2,3,4,5]
4

Step 02

Core Insight

What unlocks the optimal approach

Brute force all the possible subarrays and find the longest that satisfies the conditions.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #2760: Longest Even Odd Subarray With Threshold
class Solution {
    public int longestAlternatingSubarray(int[] nums, int threshold) {
        int ans = 0, n = nums.length;
        for (int l = 0; l < n; ++l) {
            if (nums[l] % 2 == 0 && nums[l] <= threshold) {
                int r = l + 1;
                while (r < n && nums[r] % 2 != nums[r - 1] % 2 && nums[r] <= threshold) {
                    ++r;
                }
                ans = Math.max(ans, r - l);
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #2760: Longest Even Odd Subarray With Threshold
func longestAlternatingSubarray(nums []int, threshold int) (ans int) {
	n := len(nums)
	for l := range nums {
		if nums[l]%2 == 0 && nums[l] <= threshold {
			r := l + 1
			for r < n && nums[r]%2 != nums[r-1]%2 && nums[r] <= threshold {
				r++
			}
			ans = max(ans, r-l)
		}
	}
	return
}

# Accepted solution for LeetCode #2760: Longest Even Odd Subarray With Threshold
class Solution:
    def longestAlternatingSubarray(self, nums: List[int], threshold: int) -> int:
        ans, n = 0, len(nums)
        for l in range(n):
            if nums[l] % 2 == 0 and nums[l] <= threshold:
                r = l + 1
                while r < n and nums[r] % 2 != nums[r - 1] % 2 and nums[r] <= threshold:
                    r += 1
                ans = max(ans, r - l)
        return ans

// Accepted solution for LeetCode #2760: Longest Even Odd Subarray With Threshold
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #2760: Longest Even Odd Subarray With Threshold
// class Solution {
//     public int longestAlternatingSubarray(int[] nums, int threshold) {
//         int ans = 0, n = nums.length;
//         for (int l = 0; l < n; ++l) {
//             if (nums[l] % 2 == 0 && nums[l] <= threshold) {
//                 int r = l + 1;
//                 while (r < n && nums[r] % 2 != nums[r - 1] % 2 && nums[r] <= threshold) {
//                     ++r;
//                 }
//                 ans = Math.max(ans, r - l);
//             }
//         }
//         return ans;
//     }
// }

// Accepted solution for LeetCode #2760: Longest Even Odd Subarray With Threshold
function longestAlternatingSubarray(nums: number[], threshold: number): number {
    const n = nums.length;
    let ans = 0;
    for (let l = 0; l < n; ++l) {
        if (nums[l] % 2 === 0 && nums[l] <= threshold) {
            let r = l + 1;
            while (r < n && nums[r] % 2 !== nums[r - 1] % 2 && nums[r] <= threshold) {
                ++r;
            }
            ans = Math.max(ans, r - l);
        }
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(k)

Approach Breakdown

BRUTE FORCE

O(n × k) time

O(1) space

For each starting index, scan the next k elements to compute the window aggregate. There are n−k+1 starting positions, each requiring O(k) work, giving O(n × k) total. No extra space since we recompute from scratch each time.

SLIDING WINDOW

O(n) time

O(k) space

The window expands and contracts as we scan left to right. Each element enters the window at most once and leaves at most once, giving 2n total operations = O(n). Space depends on what we track inside the window (a hash map of at most k distinct elements, or O(1) for a fixed-size window).

Shortcut: Each element enters and exits the window once → O(n) amortized, regardless of window size.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Shrinking the window only once

Wrong move: Using `if` instead of `while` leaves the window invalid for multiple iterations.

Usually fails on: Over-limit windows stay invalid and produce wrong lengths/counts.

Fix: Shrink in a `while` loop until the invariant is valid again.