Longest Continuous Increasing Subsequence

Given an unsorted array of integer 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:

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.

Approach:

C++

#include <bits/stdc++.h>
using namespace std;

int findLengthOfLCIS(vector<int> &nums)
{
    int prevCount = INT_MIN;
    int currentCount = 0;
    if (nums.size() == 0)
        return 0;
    for (int i = 0; i < nums.size() - 1; i++)
    {
        if (nums[i + 1] - nums[i] > 0)
            currentCount++;
        else
        {
            prevCount = max(prevCount, currentCount);
            currentCount = 0;
        }
    }
    return max(prevCount, currentCount) + 1;
}

int main()
{
    vector<int> nums = {1, 3, 5, 4, 7};

    cout << findLengthOfLCIS(nums);

    return 0;
}