Given an integer array nums
, return the length of the longest strictly increasing subsequence.
A subsequence is a sequence that can be derived from an array by deleting some or no elements without changing the order of the remaining elements. For example, [3,6,2,7]
is a subsequence of the array [0,3,1,6,2,2,7]
.
Example 1:
Input: nums = [10,9,2,5,3,7,101,18]
Output: 4
Explanation: The longest increasing subsequence is [2,3,7,101], therefore the length is 4.
Example 2:
Input: nums = [0,1,0,3,2,3]
Output: 4
Example 3:
Input: nums = [7,7,7,7,7,7,7]
Output: 1
Constraints:
-
1 <= nums.length <= 2500
-
-104 <= nums[i] <= 104
Follow up: Can you come up with an algorithm that runs in O(n log(n))
time complexity?
SOLUTION:
class Solution:
def lis(self, nums, i):
if i == 0:
return 1
if self.cache[i] != -1:
return self.cache[i]
mlen = 1
for j in range(i):
if nums[j] < nums[i]:
curr = 1 + self.lis(nums, j)
mlen = max(mlen, curr)
self.cache[i] = mlen
return mlen
def lengthOfLIS(self, nums: List[int]) -> int:
n = len(nums)
self.cache = [-1] * n
mlen = 1
for i in range(n):
curr = self.lis(nums, i)
mlen = max(mlen, curr)
return mlen
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