readme.md

300. Longest Increasing Subsequence

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?

Solution

class Solution:
    def lengthOfLIS(self, nums: List[int]) -> int:
        if not nums:
            return 0

        n = len(nums)
        dp = [1] * n

        for i in range(1, n):
            for j in range(i):
                if nums[j] < nums[i]:
                    dp[i] = max(dp[i], dp[j] + 1)

        return max(dp)

Thoughts

Time Complexity

O(n^2): The time complexity is quadratic because, for each element nums[i], we potentially compare it with all previous elements nums[j] where j < i. This results in roughly 1/2 * n * (n-1) comparisons in the worst case.

Space Complexity

O(n): We need a dp array of size n to store the length of the LIS ending at each index.