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//! Best Time to Buy and Sell Stock III [leetcode: best_time_to_buy_and_sell_stock_III](https://leetcode.com/problems/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). //! //! ***Example1:*** //! //! ``` //! Input: [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. //! ``` //! //! ***Example2:*** //! //! ``` //! 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. //! ``` //! //! ***Example3:*** //! //! ``` //! Input: [7,6,4,3,1] //! Output: 0 //! Explanation: In this case, no transaction is done, i.e. max profit = 0. //! ``` /// # Solutions /// /// # Approach 1: Dynamic Programming /// /// * Time complexity: O(n) /// /// * Space complexity: O(n) /// /// * Runtime: 16 ms /// * Memory: 5.7 MB /// /// ```rust /// use std::cmp; /// /// impl Solution { /// pub fn max_profit(prices: Vec<i32>) -> i32 { /// if prices.len() < 2 { return 0; } /// /// let mut result = 0; /// let n = prices.len(); /// let mut profits = vec![vec![vec![0; 2]; 3]; n]; /// profits[0][0][1] = -prices[0]; /// profits[0][1][1] = -prices[0]; /// /// for i in 1..n { /// profits[i][0][0] = profits[i - 1][0][0]; /// profits[i][0][1] = cmp::max(profits[i - 1][0][1], profits[i - 1][0][0] - prices[i]); /// /// profits[i][1][0] = cmp::max(profits[i - 1][1][0], profits[i - 1][0][1] + prices[i]); /// profits[i][1][1] = cmp::max(profits[i - 1][1][1], profits[i - 1][1][0] - prices[i]); /// /// profits[i][2][0] = cmp::max(profits[i - 1][2][0], profits[i - 1][1][1] + prices[i]); /// } /// /// result = cmp::max(profits[n-1][0][0], cmp::max(profits[n-1][1][0], profits[n-1][2][0])); /// result /// } /// } /// ``` /// /// # Approach 2: Dynamic Programming /// /// * Time complexity: O(n) /// /// * Space complexity: O(1) /// /// * Runtime: 0 ms /// * Memory: 2.6 MB /// /// ```rust /// use std::cmp::max; /// /// impl Solution { /// pub fn max_profit(prices: Vec<i32>) -> i32 { /// if prices.len() < 2 { return 0; } /// /// let kk = 2; /// let mut profits = vec![vec![0; prices.len()]; kk+1]; /// let mut max_profit = 0; /// for k in 1..=kk { /// let mut tmp_max = profits[k-1][0] - prices[0]; /// for i in 1..prices.len() { /// profits[k][i] = max(profits[k][i-1], prices[i] + tmp_max); /// tmp_max = max(tmp_max, profits[k-1][i] - prices[i]); /// max_profit = max(profits[k][i], max_profit); /// } /// } /// /// max_profit /// } /// } /// ``` /// reference: /// * [A clean DP solution which generalizes to k /// transactions](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/discuss/39608/a-clean-dp-solution-which-generalizes-to-k-transactions) /// * [Rust, generic K transactions solution with O(N) time and O(N) space, /// 中文注释](https://leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/discuss/238936/Rust-generic-K-transactions-solution-with-O(N)-time-and-O(N)-space) /// pub fn max_profit(prices: Vec<i32>) -> i32 { if prices.len() < 2 { return 0; } let kk = 2; let mut profits = vec![vec![0; prices.len()]; kk+1]; let mut max_profit = 0; for k in 1..=kk { let mut tmp_max = profits[k-1][0] - prices[0]; for i in 1..prices.len() { profits[k][i] = i32::max(profits[k][i-1], prices[i] + tmp_max); tmp_max = i32::max(tmp_max, profits[k-1][i] - prices[i]); max_profit = i32::max(profits[k][i], max_profit); } } max_profit }