# Leetcode Problem 121&122&123. Best Time to Buy and Sell Stock I/II/ III

# Problem 121:

Say you have an array for which theith element is the price of a given stock on dayi.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.

**Example 1:**

**Input:** [7,1,5,3,6,4]

**Output:** 5

**Explanation:** Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.

Not 7-1 = 6, as selling price needs to be larger than buying price.

**Example 2:**

**Input:** [7,6,4,3,1]

**Output:** 0

**Explanation:** In this case, no transaction is done, i.e. max profit = 0

# Method1: DP

dp[i] = max(dp[i -1], prices[i] — buy)

buy is always the previous lowest price.

class Solution(object):

def maxProfit(self, prices):

if not prices:

return 0 dp = [0 for __ in range(len(prices))]

buy, n = prices[0], len(prices)

for i in range(1, n):

dp[i] = max(dp[i - 1], prices[i] - buy)

buy = min(prices[i], buy)

return dp[-1]

# Method2: Without Dp

(This question can be solved directly because of its simplicity.)

class Solution(object):

def maxProfit(self, prices):

if not prices :

return 0 ans, buy, n = 0, prices[0], len(prices) for i in range(n):

ans = max(ans, prices[i]- buy)

buy = min(prices[i], buy)

return ans

# Problem 122: Best Time to Buy and Sell Stock II

Say you have an array prices for which theith element is the price of a given stock on dayi.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:You may not engage in multiple transactions at the same time (i.e., you must sell the stock before you buy again).