> ## Documentation Index
> Fetch the complete documentation index at: https://leetcode-py.wisl.dev/llms.txt
> Use this file to discover all available pages before exploring further.

> ## Agent Instructions
> leetcode-py is a Python LeetCode practice environment generator with one CLI: lcpy. It is not a service or platform.
> Each problem is a directory under leetcode/ with README.md, solution.py, test_solution.py, helpers.py, and playground.ipynb. lcpy gen creates them from JSON templates bundled with the package.
> Examples are backed by tests; copy them verbatim.

# Best Time to Buy and Sell Stock II

> Tested Python solution for LeetCode 122 with 18 pytest cases. Generate a practice environment with lcpy.

LeetCode 122, Medium. Topics: Array, Dynamic Programming, Greedy. [View on LeetCode](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-ii/description/).

Generate this problem as a practice environment: tested reference solution, 18 [parametrized pytest cases](/practice/testing), and a playground notebook:

```bash theme={"theme":{"light":"github-light","dark":"github-dark"}}
lcpy gen -n 122   # by problem number
lcpy gen -s best_time_to_buy_and_sell_stock_ii   # by problem name
```

## Problem

You are given an integer array `prices` where `prices[i]` is the price of a given stock on the `i^th` day.

On each day, you may decide to buy and/or sell the stock. You can only hold **at most one** share of the stock at any time. However, you can sell and buy the stock multiple times on the **same day**, ensuring you never hold more than one share of the stock.

Find and return *the **maximum** profit you can achieve*.

### Examples

```
Input: prices = [7,1,5,3,6,4]
Output: 7
```

**Explanation:** Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
Total profit is 4 + 3 = 7.

```
Input: prices = [1,2,3,4,5]
Output: 4
```

**Explanation:** Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.
Total profit is 4.

```
Input: prices = [7,6,4,3,1]
Output: 0
```

**Explanation:** There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0.

### Constraints

* 1 \<= prices.length \<= 3 \* 10^4
* 0 \<= prices\[i] \<= 10^4

## Solution

Reference implementation from [solution.py on GitHub](https://github.com/wislertt/leetcode-py/blob/main/leetcode/best_time_to_buy_and_sell_stock_ii/solution.py), full suite in [test\_solution.py](https://github.com/wislertt/leetcode-py/blob/main/leetcode/best_time_to_buy_and_sell_stock_ii/test_solution.py):

```python theme={"theme":{"light":"github-light","dark":"github-dark"}}
class Solution:
    # Time: O(n)
    # Space: O(1)
    def max_profit(self, prices: list[int]) -> int:
        profit = 0

        for day in range(1, len(prices)):
            # Capture every positive upswing; sum equals any multi-day strategy
            if prices[day] > prices[day - 1]:
                profit += prices[day] - prices[day - 1]

        return profit
```

## Complexity

| Time | Space |
| ---- | ----- |
| O(n) | O(1)  |

## Tags

[NeetCode 250](/catalog/neetcode-250), [NeetCode All](/catalog/neetcode).