LeetCode #3350 — MEDIUM

Adjacent Increasing Subarrays Detection II

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

Given an array nums of n integers, your task is to find the maximum value of k for which there exist two adjacent subarrays of length k each, such that both subarrays are strictly increasing. Specifically, check if there are two subarrays of length k starting at indices a and b (a < b), where:

Both subarrays nums[a..a + k - 1] and nums[b..b + k - 1] are strictly increasing.
The subarrays must be adjacent, meaning b = a + k.

Return the maximum possible value of k.

A subarray is a contiguous non-empty sequence of elements within an array.

Example 1:

Input: nums = [2,5,7,8,9,2,3,4,3,1]

Output: 3

Explanation:

The subarray starting at index 2 is [7, 8, 9], which is strictly increasing.
The subarray starting at index 5 is [2, 3, 4], which is also strictly increasing.
These two subarrays are adjacent, and 3 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist.

Example 2:

Input: nums = [1,2,3,4,4,4,4,5,6,7]

Output: 2

Explanation:

The subarray starting at index 0 is [1, 2], which is strictly increasing.
The subarray starting at index 2 is [3, 4], which is also strictly increasing.
These two subarrays are adjacent, and 2 is the maximum possible value of k for which two such adjacent strictly increasing subarrays exist.

Constraints:

2 <= nums.length <= 2 * 10⁵
-10⁹ <= nums[i] <= 10⁹

Step 01

Brute Force Baseline

Problem summary: Given an array nums of n integers, your task is to find the maximum value of k for which there exist two adjacent subarrays of length k each, such that both subarrays are strictly increasing. Specifically, check if there are two subarrays of length k starting at indices a and b (a < b), where: Both subarrays nums[a..a + k - 1] and nums[b..b + k - 1] are strictly increasing. The subarrays must be adjacent, meaning b = a + k. Return the maximum possible value of k. 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 · Binary Search

Example 1

[2,5,7,8,9,2,3,4,3,1]

Example 2

[1,2,3,4,4,4,4,5,6,7]

Step 02

Core Insight

What unlocks the optimal approach

Find the boundaries between strictly increasing subarrays.
Can we use binary search?

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 #3350: Adjacent Increasing Subarrays Detection II
class Solution {
    public int maxIncreasingSubarrays(List<Integer> nums) {
        int ans = 0, pre = 0, cur = 0;
        int n = nums.size();
        for (int i = 0; i < n; ++i) {
            ++cur;
            if (i == n - 1 || nums.get(i) >= nums.get(i + 1)) {
                ans = Math.max(ans, Math.max(cur / 2, Math.min(pre, cur)));
                pre = cur;
                cur = 0;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #3350: Adjacent Increasing Subarrays Detection II
func maxIncreasingSubarrays(nums []int) (ans int) {
	pre, cur := 0, 0
	for i, x := range nums {
		cur++
		if i == len(nums)-1 || x >= nums[i+1] {
			ans = max(ans, max(cur/2, min(pre, cur)))
			pre, cur = cur, 0
		}
	}
	return
}

# Accepted solution for LeetCode #3350: Adjacent Increasing Subarrays Detection II
class Solution:
    def maxIncreasingSubarrays(self, nums: List[int]) -> int:
        ans = pre = cur = 0
        for i, x in enumerate(nums):
            cur += 1
            if i == len(nums) - 1 or x >= nums[i + 1]:
                ans = max(ans, cur // 2, min(pre, cur))
                pre, cur = cur, 0
        return ans

// Accepted solution for LeetCode #3350: Adjacent Increasing Subarrays Detection II
impl Solution {
    pub fn max_increasing_subarrays(nums: Vec<i32>) -> i32 {
        let n = nums.len();
        let (mut ans, mut pre, mut cur) = (0, 0, 0);

        for i in 0..n {
            cur += 1;
            if i == n - 1 || nums[i] >= nums[i + 1] {
                ans = ans.max(cur / 2).max(pre.min(cur));
                pre = cur;
                cur = 0;
            }
        }

        ans
    }
}

// Accepted solution for LeetCode #3350: Adjacent Increasing Subarrays Detection II
function maxIncreasingSubarrays(nums: number[]): number {
    let [ans, pre, cur] = [0, 0, 0];
    const n = nums.length;
    for (let i = 0; i < n; ++i) {
        ++cur;
        if (i === n - 1 || nums[i] >= nums[i + 1]) {
            ans = Math.max(ans, (cur / 2) | 0, Math.min(pre, cur));
            [pre, cur] = [cur, 0];
        }
    }
    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(log n)

Space

O(1)

Approach Breakdown

LINEAR SCAN

O(n) time

O(1) space

Check every element from left to right until we find the target or exhaust the array. Each comparison is O(1), and we may visit all n elements, giving O(n). No extra space needed.

BINARY SEARCH

O(log n) time

O(1) space

Each comparison eliminates half the remaining search space. After k comparisons, the space is n/2ᵏ. We stop when the space is 1, so k = log₂ n. No extra memory needed — just two pointers (lo, hi).

Shortcut: Halving the input each step → O(log n). Works on any monotonic condition, not just sorted arrays.

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.

Boundary update without `+1` / `-1`

Wrong move: Setting `lo = mid` or `hi = mid` can stall and create an infinite loop.

Usually fails on: Two-element ranges never converge.

Fix: Use `lo = mid + 1` or `hi = mid - 1` where appropriate.