Introduction

In this blog post, we will explore how to solve the LeetCode problem "300" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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?

Explanation

Here's the breakdown of the approach, complexities, and the Python code:

• High-Level Approach:

  • Use dynamic programming with a "tails" array. The tails array stores the smallest tail of all increasing subsequences of a given length.
  • Iterate through the input nums array. For each number, either extend the longest increasing subsequence found so far (if the number is larger than the current tail) or update a tail within the tails array using binary search to maintain the smallest possible tail for each subsequence length.
  • The length of the tails array at the end represents the length of the longest increasing subsequence.

• Complexity:

  • Runtime Complexity: O(n log n)
  • Storage Complexity: O(n)

Code

    def lengthOfLIS(nums):
    """
    Finds the length of the longest strictly increasing subsequence in an array.

    Args:
        nums: A list of integers.

    Returns:
        The length of the longest strictly increasing subsequence.
    """

    tails = []  # tails[i] is the smallest tail of all increasing subsequences with length i+1

    for num in nums:
        # If num is greater than all tails, extend the longest increasing subsequence.
        if not tails or num > tails[-1]:
            tails.append(num)
        else:
            # Binary search to find the smallest tail that is greater than or equal to num.
            l, r = 0, len(tails) - 1
            while l <= r:
                mid = (l + r) // 2
                if tails[mid] < num:
                    l = mid + 1
                else:
                    r = mid - 1
            # Replace the tail with num to potentially form a smaller tail for a subsequence of the same length.
            tails[l] = num

    return len(tails)