## 1. Problem

121. Best Time to Buy and Sell Stock

You are given an array prices where prices[i] is the price of a given stock on the ith day.

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

Example 1:

```
Input: prices = [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.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.
```

Example 2:

```
Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.
```

Constraints:

- 1 <= prices.length <= 10^5
- 0 <= prices[i] <= 10^4

## 2. Solution

I solve this problem like this.

### 2.1. Time Limit Exceeded Solution

First i solve this problem like this way. But, this solution is so slow.

- Complexity
- Time complexity : O(N^2)
- Space complexity : O(N)

- Step
- By looping through the for statement, we find the value of max each time.

```
class Solution:
def maxProfit(self, prices: List[int]) -> int:
answer = 0
for idx, price in enumerate(prices[:-1]):
answer = max(max(prices[idx+1:]) - price, answer)
return answer
```

```
class Solution:
def maxProfit(self, prices: List[int]) -> int:
if len(prices) <= 1:
return 0
maxx = max(prices[1:])
answer = maxx - prices[0] if maxx - prices[0] > 0 else 0
for idx, price in enumerate(prices[1:-1]):
if price >= maxx:
maxx = max(prices[idx+1 + 1:])
answer = max(answer, maxx - price)
return answer
```

### 2.2. Make the largest of the values after the array index

This solution can solve this problem.

- Complexity
- Time complexity : O(N)
- Space complexity : O(N)

- Step
- By looping through the for statement from the back, i make a max_list.
- Check prices

```
class Solution:
def maxProfit(self, prices: List[int]) -> int:
answer = 0
if len(prices) <= 1:
return 0
max_list = [prices[-1]]
for price in prices[:-1][::-1]:
max_list.append(max(max_list[-1], price))
max_list.reverse()
for idx, price in enumerate(prices[:-1]):
answer = max(answer, max_list[idx+1] - price)
return answer
```

### 2.3. Easy Math Approach

I check other user’s solution. This is more best. Space complexity is O(1).

- Complexity
- Time complexity : O(N)
- Space complexity : O(1)

- Step
- renew buy, sell value while for statement
- price < buy : now price is lower than before buy value, so change buy price.
- price > sell : now price is bigger than before sell value, so change sell value.

- renew buy, sell value while for statement

```
class Solution:
def maxProfit(self, prices: List[int]) -> int:
buy, sell = prices[0], prices[0]
current_profit = sell - buy
for price in prices:
if price < buy:
buy = price
sell = price
if price > sell:
sell = price
if sell - buy > current_profit:
current_profit = sell - buy
return current_profit
```