#3350

Medium

auto_awesomeBinary search over the valid answer space

LeetCode Problem Workspace

Adjacent Increasing Subarrays Detection II

Find the largest k where two adjacent strictly increasing subarrays of length k exist using binary search techniques.

Problem Statement

Given an integer array nums, determine the maximum integer k such that there exist two consecutive subarrays of length k where each subarray is strictly increasing. Each subarray must be contiguous and follow the previous one without gaps.

Return the maximum possible value of k. A subarray is defined as a non-empty contiguous sequence within nums. Ensure your solution efficiently handles arrays up to 2 * 10^5 elements using the binary search pattern over the valid answer space.

Examples

Example 1

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

Output: 3

Example 2

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

Output: 2

Constraints

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

Solution Approach

Binary Search over k

Use binary search on k from 1 to n/2. For each midpoint k, check if two adjacent increasing subarrays of length k exist. Adjust search bounds based on whether a valid pair is found.

Sliding Window Validation

Maintain a window of length k and track whether each subarray is strictly increasing. Slide the window across the array to find consecutive valid subarrays efficiently.

Optimization with Precomputed Increases

Precompute the lengths of increasing sequences ending at each index. Use this information to quickly check if any subarray of length k is strictly increasing without rescanning every element.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(n log n) due to binary search on k and linear validation per candidate. Space complexity is O(n) for storing precomputed increasing lengths.

What Interviewers Usually Probe

Expect candidates to explain why naive O(n^2) checking fails for large arrays.
Look for clear reasoning on using binary search over answer space rather than array indices.
Candidate should handle edge cases where no valid adjacent increasing subarrays exist.

Common Pitfalls or Variants

Common pitfalls

Off-by-one errors when checking subarray boundaries.
Failing to handle overlapping subarrays correctly.
Not precomputing increasing lengths, leading to timeout on large inputs.

Follow-up variants

Find the maximum k for two non-adjacent strictly increasing subarrays.
Determine maximum k for decreasing subarrays instead of increasing.
Return all starting indices of valid adjacent increasing subarrays of maximum length.

FAQ

What is the main approach for Adjacent Increasing Subarrays Detection II?

Use binary search over possible lengths k and validate each with a sliding window or precomputed increasing sequences.

How do I efficiently check if a subarray is strictly increasing?

Precompute the lengths of increasing sequences ending at each index to avoid rescanning elements.

What if no adjacent increasing subarrays exist?

Return 0 since no valid k satisfies the problem condition.

Can overlapping subarrays be considered adjacent?

No, adjacent subarrays must be consecutive and non-overlapping.

Why binary search works for this problem?

Because the set of valid k values is monotonic: if k is valid, all smaller lengths are also valid.

terminal

Solution

Solution 1: Single Pass

We can use a single pass to calculate the maximum length of adjacent increasing subarrays $\textit{ans}$. Specifically, we maintain three variables: $\textit{cur}$ and $\textit{pre}$ represent the length of the current increasing subarray and the previous increasing subarray respectively, while $\textit{ans}$ represents the maximum length of adjacent increasing subarrays.

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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

Continue Practicing

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