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:
nums[a..a + k - 1] and nums[b..b + k - 1] are strictly increasing.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:
2 is [7, 8, 9], which is strictly increasing.5 is [2, 3, 4], which is also strictly increasing.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] <= 1000This approach uses a single loop and checks both subarrays simultaneously to see if they are strictly increasing. By maintaining two pointers, one for the first subarray and one for the potential adjacent subarray, we can efficiently determine if there exists any pair of strictly increasing adjacent subarrays.
This function first checks if a subarray starting at the given index with length k is strictly increasing. The main function iterates through the array with initial index such a way that enough elements remain for two subarrays of length k. If at any such index both subarrays are strictly increasing, it returns true.
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Time Complexity: O(n * k), where n is the length of the array. The two pointer approach still checks each element of both subarrays.
Space Complexity: O(1), as no additional space is used apart from a few counters.
The sliding window technique can provide a more efficient approach by iterating through the array once and maintaining running comparisons. Continuously evaluate if the latest element extends an increasing pattern and reset boundaries where necessary.
The solution uses a counter that increments through every consecutive pair increase. Two adjacent subarrays need k-1 increases each plus the adjacent element, hence k*2-1 successful increments confirm the presence of two such subarrays.
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Time Complexity: O(n)
Space Complexity: O(1)
| Approach | Complexity |
|---|---|
| Using Two Pointers for Check | Time Complexity: O(n * k), where n is the length of the array. The two pointer approach still checks each element of both subarrays. |
| Using Sliding Window Technique | Time Complexity: O(n) |
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