674. Longest Continuous Increasing Subsequence

Problem Statement

link: LeetCode.cn LeetCode 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:

Input: nums = [1,3,5,4,7] Output: 3

Input: nums = [2,2,2,2,2] Output: 1

Constraints: 1 <= nums.length <= 104 -109 <= nums[i] <= 109

Solution Approach

The solution iteratively expands the right boundary of a sliding window while tracking the longest continuous increasing subsequence, resetting the left boundary whenever the sequence stops increasing.

Algorithm

1. Initialize Pointers and Variables: Start with two pointers, l and r, both at the beginning of the array, and initialize a variable res to keep track of the maximum length of any increasing subsequence found.
2. Expand the Right Pointer: Move the right pointer r forward while the sequence is strictly increasing (nums[r + 1] > nums[r]). Update the maximum length res when the sequence ends.
3. Reset Pointers: Once the sequence stops increasing, move both pointers to the next position (l = r + 1) and repeat the process until the entire array is scanned. Return the maximum length res as the result.

Implement

    class Solution:
  def findLengthOfLCIS(self, nums: List[int]) -> int:
      res = 0
      n = len(nums) 
      if not n:
          return 0
      l, r = 0, 0 
      if n == 1:  
          return 1
      while r < n - 1: 
          while r < n - 1 and nums[r + 1] > nums[r]: 
              r += 1 
          res = max(res, r - l + 1) 
          l = r = r + 1  
      return res  

    
    

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