# leetcode\_122

## 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:

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

```
Example 1:

Input: [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.
Example 2:

Input: [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.
             Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are
             engaging multiple transactions at the same time. You must sell before buying again.
Example 3:

Input: [7,6,4,3,1]
Output: 0
Explanation: In this case, no transaction is done, i.e. max profit = 0.
```

## Solutions

1. **Straight forward**
2. To achieve the maximum profit, we need to buy at the start of each ascending slope and sell at the end the slope.
3. This process equals to summing up all the price difference between two consecutive days.

```cpp
class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.size() < 2) return 0;
        int sum = 0;
        for (int i = 1; i < prices.size(); i++) {
            if (prices[i] > prices[i - 1])
                sum += prices[i] - prices[i - 1];
        }

        return sum;
    }
};
```

1. **Dynamic programming**

See problem 123.