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:
nums[a..a + k - 1] and nums[b..b + k - 1] are strictly increasing.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:
[7, 8, 9], which is strictly increasing.[2, 3, 4], which is also strictly increasing.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:
[1, 2], which is strictly increasing.[3, 4], which is also strictly increasing.k for which two such adjacent strictly increasing subarrays exist.
Constraints:
2 <= nums.length <= 2 * 105-109 <= nums[i] <= 109The idea is to use a dynamic programming technique that calculates the length of the longest increasing subarray that ends at each position.
We maintain an array inc[], where inc[i] stores the length of the longest strictly increasing subarray ending at index i.
Once this array is filled, the solution is to find two adjacent values in inc[] where both values are at least k. Because they are adjacent, they indicate two subarrays of length k each, which are strictly increasing.
The C solution uses an array inc to track the length of increasing subarrays. It iterates through the input array, updating inc to hold the length of the current increasing sequence ending at each position. Finally, it looks for two adjacent lengths in inc that are both larger than max_k.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n) due to a single pass through the array to fill the inc array.
Space Complexity: O(n) due to the inc array.
This method optimizes the space usage by relying on two variables instead of an entire array. The goal is to calculate the length of current increasing sequences directly within the loop.
We keep two lengths, current and previous, to track increasing subarrays as we iterate through the input array.
The C solution uses two variables instead of an array to track the lengths of the current and previous increasing subarrays, reducing the space complexity.
C++
Java
Python
C#
JavaScript
Time Complexity: O(n).
Space Complexity: O(1).
| Approach | Complexity |
|---|---|
| Approach 1: Dynamic Programming | Time Complexity: O(n) due to a single pass through the array to fill the |
| Approach 2: Optimized Single Pass | Time Complexity: O(n). |
LeetCode was HARD until I Learned these 15 Patterns • Ashish Pratap Singh • 1,002,307 views views
Watch 9 more video solutions →Practice Adjacent Increasing Subarrays Detection II with our built-in code editor and test cases.
Practice on FleetCode