# 300 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?

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 `````` ``````''' DP, O(N^2) ''' class Solution: def lengthOfLIS(self, nums: List[int]) -> int: n = len(nums) dp = [1] * n for i in range(n): for j in range(i+1, n): if nums[j] > nums[i]: dp[j] = max(dp[j], dp[i] + 1) return max(dp) ''' Binary Search, O(Nâ‹…log(N)) ''' class Solution: def lengthOfLIS(self, nums: List[int]) -> int: sub = [] for num in nums: i = bisect_left(sub, num) # If num is greater than any element in sub if i == len(sub): sub.append(num) # Otherwise, replace the first element in sub greater than or equal to num else: sub[i] = num return len(sub)``````