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This approach involves iterating through the array while maintaining a variable to track the length of the current continuous increasing subsequence. As we encounter each element, we check if it is larger than the previous one to decide if we should continue the current subsequence or start a new one. We also maintain a global maximum to store the maximum length found during our iteration.
Time Complexity: O(n), where n is the number of elements in the array.
Space Complexity: O(1), as we use a constant amount of space.
1def find_length_of_LCIS(nums):
2 if not nums:
3 return 0
4 max_len = 1
5 curr_len = 1
6
7 for i in range(1, len(nums)):
8 if nums[i] > nums[i - 1]:
9 curr_len += 1
10 max_len = max(max_len, curr_len)
11 else:
12 curr_len = 1
13
14 return max_len
15
16nums = [1, 3, 5, 4, 7]
17print(find_length_of_LCIS(nums))In the Python solution, we loop through the list of numbers, check if the sequence continues to be increasing, and update or reset the current sequence length accordingly.
The greedy approach leverages a single pass through the array to determine the maximum length of an increasing subsequence. This method is optimal because it immediately processes each element only once, updating the sequence lengths on the fly, and computing the maximum without revisiting any part of the array.
Time Complexity: O(n), where n is the number of elements in the array.
Space Complexity: O(1), as no additional space other than variables is utilized.
This code utilizes a greedy solution where it enters the loop with the ability to modify the beginning of sequences implicitly by re-syncing the currLen, ensuring we keep the cost constant per iteration.