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?

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'''
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(N⋅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)