123. Best Time to Buy and Sell Stock III

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

Example 4:

Input: prices = [1]
Output: 0

 

Constraints:

  • 1 <= prices.length <= 105
  • 0 <= prices[i] <= 105

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);
}

Having problems with this solution? Click here to submit an issue on github.