#3349

Easy

auto_awesomeArray-driven solution strategy

LeetCode Problem Workspace

Adjacent Increasing Subarrays Detection I

Check if an array contains two adjacent strictly increasing subarrays of length k using an array-driven approach efficiently.

array

Problem Statement

Given an array nums of n integers and an integer k, determine whether two adjacent subarrays of length k exist such that both subarrays are strictly increasing. Adjacent means the first subarray ends immediately before the second starts, and each subarray contains elements in strictly increasing order.

Return true if such two adjacent increasing subarrays exist; otherwise, return false. For example, given nums = [2,5,7,8,9,2,3,4,3,1] with k = 3, the output should be true because [5,7,8] and [7,8,9] are consecutive increasing subarrays of length 3. If no such pairs exist, return false.

Examples

Example 1

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

Output: true

Example 2

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

Output: false

Example details omitted.

Constraints

2 <= nums.length <= 100
1 < 2 * k <= nums.length
-1000 <= nums[i] <= 1000

Solution Approach

Scan and Track Increasing Lengths

Iterate through nums and maintain a counter for the length of the current strictly increasing sequence. Reset the counter when a non-increasing element is encountered. This allows identification of all subarrays of length k without nested loops.

Record Valid Subarray Indices

Whenever an increasing sequence reaches length k, record its starting index. Maintain a list of these indices and check for adjacent pairs where the end of one subarray is immediately before the start of another.

Check Adjacency for Result

Compare consecutive recorded subarray indices to see if any two are exactly k elements apart. If such a pair exists, return true. Otherwise, return false after the scan completes.

Complexity Analysis

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

Time complexity is O(n) because each element is processed once while tracking increasing sequences. Space complexity is O(n) in the worst case to store valid subarray starting indices, but can be reduced by checking adjacency on the fly.

What Interviewers Usually Probe

Expecting array traversal without nested loops
Looking for handling of overlapping subarrays correctly
Checking for attention to edge cases like k near n/2

Common Pitfalls or Variants

Common pitfalls

Confusing non-strictly increasing sequences as valid
Not handling overlapping subarrays correctly
Using nested loops leading to unnecessary O(n*k) complexity

Follow-up variants

Detect adjacent decreasing subarrays of length k
Find three consecutive increasing subarrays
Check for subarrays with at least k increasing elements

FAQ

What is the pattern behind Adjacent Increasing Subarrays Detection I?

It focuses on detecting two consecutive subarrays of length k where each subarray is strictly increasing, leveraging array scanning.

Can k equal half of nums length?

Yes, as long as 1 < 2*k <= nums.length, the algorithm correctly handles edge cases.

How do overlapping subarrays affect detection?

Overlapping is allowed only if the subarrays are strictly adjacent; GhostInterview ensures adjacency is checked precisely.

What is the time complexity of this solution?

It is O(n) because each element is scanned once to track increasing sequences and record valid subarrays.

Does this problem require extra space for tracking?

You can store starting indices of increasing subarrays, but checking adjacency on the fly can reduce space usage.

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Solution

Solution 1: Single Pass

According to the problem description, we only need to find the maximum length of adjacent increasing subarrays $\textit{mx}$. If $\textit{mx} \ge k$, then there exist two adjacent strictly increasing subarrays of length $k$.

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class Solution:
    def hasIncreasingSubarrays(self, nums: List[int], k: int) -> bool:
        mx = pre = cur = 0
        for i, x in enumerate(nums):
            cur += 1
            if i == len(nums) - 1 or x >= nums[i + 1]:
                mx = max(mx, cur // 2, min(pre, cur))
                pre, cur = cur, 0
        return mx >= k

Continue Practicing

#3350 adjacent increasing subarrays detection ii #3347 maximum frequency of an element after performing operations ii #3351 sum of good subsequences