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longest_increasing_subsequence_o_nlogn.py
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longest_increasing_subsequence_o_nlogn.py
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#############################
# Author: Aravind Kashyap
# File: lis.py
# comments: This programme outputs the Longest Strictly Increasing Subsequence in
# O(NLogN) Where N is the Number of elements in the list
#############################
from __future__ import annotations
def ceil_index(v, left, right, key):
while right - left > 1:
middle = (left + right) // 2
if v[middle] >= key:
right = middle
else:
left = middle
return right
def longest_increasing_subsequence_length(v: list[int]) -> int:
"""
>>> longest_increasing_subsequence_length([2, 5, 3, 7, 11, 8, 10, 13, 6])
6
>>> longest_increasing_subsequence_length([])
0
>>> longest_increasing_subsequence_length([0, 8, 4, 12, 2, 10, 6, 14, 1, 9, 5, 13,
... 3, 11, 7, 15])
6
>>> longest_increasing_subsequence_length([5, 4, 3, 2, 1])
1
"""
if len(v) == 0:
return 0
tail = [0] * len(v)
length = 1
tail[0] = v[0]
for i in range(1, len(v)):
if v[i] < tail[0]:
tail[0] = v[i]
elif v[i] > tail[length - 1]:
tail[length] = v[i]
length += 1
else:
tail[ceil_index(tail, -1, length - 1, v[i])] = v[i]
return length
if __name__ == "__main__":
import doctest
doctest.testmod()