Longest Increasing Subsequence

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