Monthly Archives: December 2019

How to buy and sell stock at the best time

Posted on by
Reply
  • Best Time to Buy and Sell Stock I
  • Best Time to Buy and Sell Stock II
  • Best Time to Buy and Sell Stock III
  • Best Time to Buy and Sell Stock IV
  • Best Time to Buy and Sell Stock with cooldown
  • Best Time to Buy and Sell Stock with transaction fee

Best Time to Buy and Sell Stock I
Say you have an array for which the ith element is the price of a given stock on day i.

If you were only permitted to complete at most one transaction (i.e., buy one and sell one share of the stock), design an algorithm to find the maximum profit.

Note that you cannot sell a stock before you buy one.

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        buy, sell = -0x7777777, 0

        for price in prices:
            buy = max(buy, -price)
            sell = max(sell, buy + price)

        return sell

Best Time to Buy and Sell Stock II
Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (i.e., buy one and sell one share of the stock multiple times).

Note that you may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        profit, buy = 0, 0x7777777

        for price in prices:
            buy = min(buy, price)

            tmp = price - buy
            
            it tmp > 0:
                profit += tmp
                buy = price

        return profit

Best Time to Buy and Sell Stock III
Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most two transactions.

Note that you may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).

class Solution:
    def maxProfit(self, prices: List[int]) -> int:
        buy_1, buy_2, sell_1, sell_2 = -0x7777777, -0x7777777, 0, 0

        for price in prices:
            buy_1 = max(buy_1, -price)
            sell_1 = max(sell_1, buy_1 + price)

            buy_2 = max(buy_2, sell_1 - price)
            sell_2 = max(sell_2, buy_2 + price)

        return sell_2

Best Time to Buy and Sell Stock IV
Say you have an array for which the i-th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most k transactions.

Note that you may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

class Solution:
    def maxProfit(self, k: int, prices: List[int]) -> int:
        if k > len(prices) >> 1:
            return sum(prices[i+1] - prices[i] for i in range(len(prices) - 1) if prices[i+1] > prices[i])
        
        cash, asset = [-0x7777777] * (k + 1), [0] * (k + 1)

        for price in prices:
            for i in range(1, k+1):
                cash[i] = max(cash[i], sell[i-1] - price)
                asset[i] = max(asset[i], cash[i] + price)

        return asset[k]

Best Time to Buy and Sell Stock with Cooldown

Say you have an array for which the ith element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete as many transactions as you like (ie, buy one and sell one share of the stock multiple times) with the following restrictions:

  • You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).
  • After you sell your stock, you cannot buy stock on the next day. (ie, cooldown 1 day)
class Solution:
    def maxProfit(self, prices: List[int]) -> int: 
        if not prices:
            return 0
        
        sell, buy, prev_sell, prev_buy = 0, -prices[0], 0, 0
        
        for price in prices:
            prev_buy = buy
            buy = max(prev_buy, prev_sell - price)
            prev_sell = sell
            sell = max(prev_sell, prev_buy + price)
            
        return sell

Best Time to Buy and Sell Stock with Transaction Fee
You are given an array of integers prices, for which the i-th element is the price of a given stock on day i; and a non-negative integer fee representing a transaction fee.

You may complete as many transactions as you like, but you need to pay the transaction fee for each transaction. You may not buy more than 1 share of a stock at a time (ie. you must sell the stock share before you buy again.)

Return the maximum profit you can make.

class Solution:
    def maxProfit(self, prices: List[int], fee: int) -> int:
        sell, buy = 0, -prices[0]
        
        for p in prices[1:]:
            sell = max(sell, buy + p - fee)
            buy = max(buy, sell - p)
            
        return sell

Verily Phone Screen Interview

Posted on by
Reply

That is a sad story…

Just several days ago, I received an email from Verily, an Alphabet company which delicates in life science. Their HR passed my application and going to move me to the phone interview. However, I messed it up…

I have to say that interview is not a difficult one. The question is like a medium level question at Leetcode:

Given a 1-dimensional axis, a man can move left or right in each time unit. How many possibilities that the man stands on x point after t time units?

I stupidly tried DP at first and struggled in how to implement the state transform formula, that wastes a lot of time.

Today, I reviewed this question and found a fairly easy solution. We do not even need Dynamic Programming. 

l + r = t   (1)
r - l = x   (2)

Once we solve this equation, we can directly calculate the number of combinations. For example, the total possibilities of that man stand on point 5 after 9 time units is C72 = 21.

Word Break

Posted on by
Reply

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

  • The same word in the dictionary may be reused multiple times in the segmentation.
  • You may assume the dictionary does not contain duplicate words.

Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
             Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false

Recently, I am going to pick up my leetcode skills.

This problem that I still remember It token me more than two days to consider, but this time it was aced in 5 minutes, as well as just in nearly 1 line core code.

It seems like practising is really useful!

from functools import lru_cache

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        @lru_cache(None)
        def dfs(s):
            return True if not s else any(dfs(s[len(word):]) for word in wordDict if s.startswith(word))
            
        return dfs(s)