Longest Continuous Increasing Subsequence

Problem

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

Solution

/**
 * @param {number[]} nums
 * @return {number}
 */
var findLengthOfLCIS = function(nums) {
    if (!nums.length) {
        return 0;
    }
    let max = 1;
    let dp = Array(nums.length).fill(1);
    for (let i = 1; i < nums.length; ++i) {
        if (nums[i - 1] < nums[i]) {
            dp[i] = dp[i - 1] + 1;
            max = Math.max(dp[i], max);
        }
    }
    return max;
};

We will implement a DP solution. Let dp[i] be the longest continuous subsequence ending and including nums[i].

• For the base case, if i = 0, then dp[0] = 1, since subsequence ends at the first element of nums.
• For the recursive case, if the current value nums[i] is greater than the previous value nums[i - 1], the longest continuous subsequence ending at the current value is the previous longest subsequence plus 1; otherwise the subsequence will only contain nums[i], which has length 1.