674.longest-continuous-increasing-subsequence.cpp

//  Tag: Array
//  Time: O(N)
//  Space: O(1)
//  Ref: -
//  Note: -

//  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
//  
//  

class Solution {
public:
    int findLengthOfLCIS(vector<int>& nums) {
        int n = nums.size();
        vector<int> dp(n , 1);

        for (int i = 1; i < n; i++) {
            if (nums[i] > nums[i - 1]) {
                dp[i] = dp[i - 1] + 1;
            }
        }

        return *max_element(dp.begin(), dp.end());
    }
};

class Solution {
public:
    int findLengthOfLCIS(vector<int>& nums) {
        int n = nums.size();
        int cur = 1;
        int res = 1;

        for (int i = 1; i < n; i++) {
            if (nums[i] > nums[i - 1]) {
                cur += 1;
            } else {
                cur = 1;
            }
            res = max(res, cur);
        }

        return res;
    }
};

class Solution {
public:
    int findLengthOfLCIS(vector<int>& nums) {
        int res = 1;
        int i = 0;
        for (auto j = 1; j < nums.size(); ++j) {
            if (nums[j] <= nums[j - 1]) {
                i = j;
            }
            res = max(res, j - i + 1);
        }
        
        return res;
    }
};