leetcode-dp

300. Longest Increasing Subsequence.py

class Solution:
    """    
    Problem Clarification
        -We must ask interviewer to go through the examples.
        -You can even go through another examples to make sure there's no misunderstanding.
        -Ask edge case, input and output data structure, sorted or not sorted, Positive or Negative?
            duplicates?
            strictly increasing?
            
    Thought Process
        If we traverse the given nums, whether current subsequence is a valid depends on whether previous subsequence is valid. So, there must be lots of overlapping subproblems.
        
        Let's define subproblems dp[i]: length of increasing subsequence end with ith element
        
        So, our goal is return max(dp[0], dp[1],.., dp[n-1]) instead of dp[n-1]
        
        base case, i == 0
            dp[0] = 1
            
        if cuurent subproblems relies on all previoust subproblems.
             if nums[i] > nums[j]:
                dp[i] = max(dp[i], dp[j] + 1)
                u don't know which subproblem dp[j] will contribute dp[i] the max lengh of increasing subsequence
                so you take maximum
        [1,2,3,4,5]
           i
         j
             i
         j j
    Note:
    1. To current subproblems i, every previous subrproblem dp[j] are already solved. 
    """
    def lengthOfLIS(self, nums: List[int]) -> int:
        """
        Don't forget to simply describe brute force solution orally during the interview
        before getting started to use any algorithm technique.
        
        Brute Force:
            The simplest approach is to try to find all increasing subsequences and then returning the maximum length of longest increasing subsequence.
        And, the time complexity is (2**n).
        However, it's inefficient, and we found the main problem could be divided into overlapping supproblem. So, we use DP.
        
        Note:
            1. Each cell in dp represent longest length of increasing subsequence considering the current element
            2. dp[i]= represents length of LIS including nums[j].
        
        T: O(n**2) because of two loops of n there
        S: O(n) because we use extra space for the memorization table.
        """
        n = len(nums)
        if n == 0:
            return 0
        if n == 1:
            return 1
        dp = [1 for _ in range(n)] # Step1: base case: dp[0] = 1
        
        for i in range(1, n):
            for j in range(i):
                # Step2: recursive formulation
                if nums[i] > nums[j]:
                    dp[i] = max(dp[j]+1, dp[i])
        return max(dp)