Longest Increasing Subsequence Python Solution

Tested Python solution for LeetCode 300 with 15 pytest cases. Generate a practice environment with lcpy.

LeetCode 300, Medium. Topics: Array, Binary Search, Dynamic Programming. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 15 parametrized pytest cases, and a playground notebook:

lcpy gen -n 300   # by problem number
lcpy gen -s longest_increasing_subsequence   # by problem name

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Problem

Given an integer array nums, return the length of the longest strictly increasing subsequence.

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Examples

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

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Constraints

• 1 <= nums.length <= 2500
• -10^4 <= nums[i] <= 10^4

Follow up: Can you come up with an algorithm that runs in O(n log(n)) time complexity?

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(n log n)
    # Space: O(n)
    def length_of_lis(self, nums: list[int]) -> int:
        """
        Binary Search + DP: tails[i] = smallest tail of all LIS of length i+1

        Example with middle replacement: [1,5,3,7,2,6,4,8]

        Step | num | tails        | Action
        -----|-----|--------------|------------------
          1  |  1  | [1]          | append
          2  |  5  | [1,5]        | append
          3  |  3  | [1,3]        | replace 5 (pos=1)
          4  |  7  | [1,3,7]      | append
          5  |  2  | [1,2,7]      | replace 3 (pos=1)
          6  |  6  | [1,2,6]      | replace 7 (pos=2)
          7  |  4  | [1,2,4]      | replace 6 (pos=2)
          8  |  8  | [1,2,4,8]    | append

        Result: len(tails) = 4
        """
        import bisect

        tails: list[int] = []

        for num in nums:
            pos = bisect.bisect_left(tails, num)
            if pos == len(tails):
                tails.append(num)
            else:
                tails[pos] = num

        return len(tails)

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Complexity

Time	Space
O(n log n)	O(n)

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Tags

Grind, Blind 75, NeetCode 150, NeetCode 250, NeetCode All, AlgoMaster 75.