Longest increasing subsequence

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        dp = [1] * len(nums)
        dp[0] = 1
        for i in range(1, len(nums)):
            for j in range(i):
                if nums[i] > nums[j]:
                    dp[i] = max(dp[i], dp[j] + 1)
        return max(dp)

Longest Increasing Subsequence

Difficulty: Medium

Given an integer array nums, return the length of the longest strictly increasing subsequence.

 

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?