Adjacent Increasing Subarrays Detection I - Video Solutions
Adjacent Increasing Subarrays Detection I | Simple and Intuitive | Leetcode 3349 | codestorywithMIK
Adjacent Increasing Subarrays Detection I - Video Solution
Watch 10 video solutions for Adjacent Increasing Subarrays Detection I, a easy level problem involving Array. This walkthrough by codestorywithMIK has 8,038 views views. Want to try solving it yourself? Practice on FleetCode or read the detailed text solution.
Problem Statement
Given an array nums of n integers and an integer k, determine whether there exist two adjacent subarrays of length k such that both subarrays are strictly increasing. Specifically, check if there are two subarrays starting at indices a and b (a < b), where:
- Both subarrays
nums[a..a + k - 1]andnums[b..b + k - 1]are strictly increasing. - The subarrays must be adjacent, meaning
b = a + k.
Return true if it is possible to find two such subarrays, and false otherwise.
Example 1:
Input: nums = [2,5,7,8,9,2,3,4,3,1], k = 3
Output: true
Explanation:
- The subarray starting at index
2is[7, 8, 9], which is strictly increasing. - The subarray starting at index
5is[2, 3, 4], which is also strictly increasing. - These two subarrays are adjacent, so the result is
true.
Example 2:
Input: nums = [1,2,3,4,4,4,4,5,6,7], k = 5
Output: false
Constraints:
2 <= nums.length <= 1001 < 2 * k <= nums.length-1000 <= nums[i] <= 1000
Approach Overview
Problem Overview: You are given an integer array and a value k. The task is to determine whether the array contains two adjacent strictly increasing subarrays, each of length k. In other words, check if there exists an index i where nums[i..i+k-1] and nums[i+k..i+2k-1] are both strictly increasing sequences.
Approach 1: Two Pointers for Check (O(n) time, O(1) space)
This method scans the array and measures the length of strictly increasing segments using two pointers. Start from the left and expand the right pointer while nums[r] > nums[r-1]. This gives the length of a continuous increasing run. If a run length is at least 2k, or if two consecutive runs each have length at least k, you can form two adjacent increasing subarrays. The algorithm only tracks boundaries and lengths, so no extra data structures are required.
The key insight: strictly increasing subarrays must lie inside increasing runs. Instead of checking every possible pair explicitly, compute run lengths and verify whether the required adjacent segments exist.
Approach 2: Sliding Window Technique (O(n) time, O(1) space)
The sliding window approach checks windows of size k while maintaining whether each window is strictly increasing. Move a window across the array and track violations where nums[i] <= nums[i-1]. Two consecutive windows are valid if both satisfy the increasing condition and the second starts exactly after the first.
To optimize, maintain a running count of increasing comparisons inside the window. As the window slides forward, update the count instead of recomputing the entire window. This converts a naive O(nk) check into an O(n) scan while keeping constant space.
This technique is a common pattern in sliding window problems where local order constraints must be verified efficiently.
Recommended for interviews: The two‑pointer run-length scan is typically what interviewers expect. It directly models how increasing sequences behave and produces a clean O(n) solution with constant memory. A sliding window implementation also works and demonstrates familiarity with two pointers and window maintenance patterns. Showing the brute reasoning first (checking segments) and then optimizing to a single pass highlights strong problem-solving skills.
Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Two Pointers (Run Length Scan) | O(n) | O(1) | Best general solution. Efficient single pass when checking increasing runs. |
| Sliding Window Technique | O(n) | O(1) | Useful when explicitly validating fixed-size windows or practicing window-based patterns. |