## 123. Best Time to Buy and Sell Stock III

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 *two* transactions.

**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:prices = [3,3,5,0,0,3,1,4]Output:6Explanation:Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3. Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

**Example 2:**

Input:prices = [1,2,3,4,5]Output:4Explanation: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:prices = [7,6,4,3,1]Output:0Explanation:In this case, no transaction is done, i.e. max profit = 0.

**Example 4:**

Input:prices = [1]Output:0

**Constraints:**

`1 <= prices.length <= 10`

^{5}`0 <= prices[i] <= 10`

^{5}

## Rust Solution

```
struct Solution;
impl Solution {
fn max_profit(prices: Vec<i32>) -> i32 {
let mut t1_cost = std::i32::MAX;
let mut t2_cost = std::i32::MAX;
let mut t1_profit = 0;
let mut t2_profit = 0;
for x in prices {
t1_cost = t1_cost.min(x);
t1_profit = t1_profit.max(x - t1_cost);
t2_cost = t2_cost.min(x - t1_profit);
t2_profit = t2_profit.max(x - t2_cost);
}
t2_profit
}
}
#[test]
fn test() {
let prices = vec![3, 3, 5, 0, 0, 3, 1, 4];
let res = 6;
assert_eq!(Solution::max_profit(prices), res);
let prices = vec![1, 2, 3, 4, 5];
let res = 4;
assert_eq!(Solution::max_profit(prices), res);
let prices = vec![7, 6, 4, 3, 1];
let res = 0;
assert_eq!(Solution::max_profit(prices), res);
}
```

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