Given an unsorted array of integers nums, return the length of the longest continuous increasing subsequence (i.e. subarray). The subsequence must be strictly increasing.
A continuous increasing subsequence is defined by two indices l and r (l < r) such that it is [nums[l], nums[l + 1], ..., nums[r - 1], nums[r]] and for each l <= i < r, nums[i] < nums[i + 1].
Example 1:
Input: nums = [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1,3,5] with length 3.
Even though [1,3,5,7] is an increasing subsequence, it is not continuous as elements 5 and 7 are separated by element
4.
Example 2:
Input: nums = [2,2,2,2,2]
Output: 1
Explanation: The longest continuous increasing subsequence is [2] with length 1. Note that it must be strictly
increasing.
Constraints:
1 <= nums.length <= 104
-109 <= nums[i] <= 109
Solutions
Solution 1: One-pass Scan
We can traverse the array $nums$, using a variable $cnt$ to record the length of the current consecutive increasing sequence. Initially, $cnt = 1$.
Then, we start from index $i = 1$ and traverse the array $nums$ to the right. Each time we traverse, if $nums[i - 1] < nums[i]$, it means that the current element can be added to the consecutive increasing sequence, so we set $cnt = cnt + 1$, and then update the answer to $ans = \max(ans, cnt)$. Otherwise, it means that the current element cannot be added to the consecutive increasing sequence, so we set $cnt = 1$.
After the traversal ends, we return the answer $ans$.
The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.
We can also use two pointers $i$ and $j$ to find each consecutive increasing sequence, and find the length of the longest consecutive increasing sequence as the answer.
The time complexity is $O(n)$, where $n$ is the length of the array $nums$. The space complexity is $O(1)$.