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
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.
Input: nums = [0,1,0,3,2,3] Output: 4
Input: nums = [7,7,7,7,7,7,7] Output: 1
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?
Use dynamic programming to solve this problem. Each position’s longest increasing subsequence is the from a previous position which is less than the position and has the largest longest increasing subsequence.
class Solution: def lengthOfLIS(self, nums: List[int]) -> int: if not nums: return 0 dp =  * len(nums) for i in range(len(nums)): for j in range(i): if nums[j] < nums[i]: dp[i] = max(dp[i], dp[j] + 1) return max(dp)
- Time complexity: O(N^2)
- Space complexity: O(N)