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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 as the largest subsequence where the consecutive elements are strictly increasing.

Example:

Input: nums = [1,3,5,4,7]
Output: 3
Explanation: The longest continuous increasing subsequence is [1, 3, 5] with a length of 3.

Clarifying Questions:

  1. Are all elements in the input array integers?
    • Yes.
  2. Can the input array contain negative numbers or zeros?
    • Yes.
  3. What is the maximum length of the input array?
    • The maximum length of the input array can be up to 10^4.
  4. Are there any constraints on the value of individual integers in the array?
    • No specific constraints, integers can be of any value within standard integer limits.

Code:

def findLengthOfLCIS(nums):
    if not nums:
        return 0
    
    # Initialize the count and max length
    max_length = 1
    current_length = 1
    
    # Iterate through the array
    for i in range(1, nums.length):
        # Check if the current element is greater than the previous element
        if nums[i] > nums[i - 1]:
            current_length += 1  # Increment current length
        else:
            max_length = max(max_length, current_length)
            current_length = 1  # Reset current length
    
    # After the loop ends, we must check once more to update the max_length.
    max_length = max(max_length, current_length)
    
    return max_length

Strategy:

  1. Initialize Variables:
    • max_length to track the maximum length of continuous increasing subsequence found.
    • current_length to track the length of the current continuous increasing subsequence.
  2. Iterate Through the Array:
    • Start iterating from the second element (index 1).
    • If the current element is greater than the previous element, increment current_length.
    • If the current element is not greater than the previous element, update max_length if current_length is greater, and reset current_length to 1.
  3. Final Comparison:
    • After the loop, there might be a case where the longest subarray ends at the last element. Hence, perform a final update check for max_length.

Time Complexity:

This implementation ensures that we’re efficiently finding the length of the longest continuous increasing subsequence with optimal time and space complexity.

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