#### Longest Increasing Subsequence O(nlogn)

p
```#############################
# 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, l, r, key):  # noqa: E741
while r - l > 1:
m = (l + r) // 2
if v[m] >= key:
r = m
else:
l = m  # noqa: E741
return r

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 =  * len(v)
length = 1

tail = v

for i in range(1, len(v)):
if v[i] < tail:
tail = 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()
```  