'Best Time to Buy and Sell Stock III' soln

Author: joewnga

leetcode/best_time_to_buy_and_sell_stock_iii.rs

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+/// 
+/// Problem: Best Time to Buy and Sell Stock III
+/// 
+/// You are given an array prices where prices[i] is the price of a given stock on the ith day.
+/// 
+/// Find the maximum profit you can achieve. You may complete at most two transactions.
+/// 
+/// Note: You may not engage in multiple transactions simultaneously (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.
+/// 
+/// Constraints:
+/// 1 <= prices.length <= 10^5
+/// 0 <= prices[i] <= 10^5
+/// 
+
+// # Solution: 
+// Time complexity: O(n) 
+// Space complexity: O(1)
+
+
+impl Solution {
+    pub fn max_profit(prices: Vec<i32>) -> i32 {
+        if prices.is_empty() {
+           return 0;
+       }
+       
+       let mut buy1 = -prices[0];  
+       let mut sell1 = 0;          
+       let mut buy2 = -prices[0];  
+       let mut sell2 = 0;          
+       
+       for i in 1..prices.len() {
+           
+           buy1 = buy1.max(-prices[i]);
+           
+           sell1 = sell1.max(buy1 + prices[i]);
+           
+           buy2 = buy2.max(sell1 - prices[i]);
+           
+           sell2 = sell2.max(buy2 + prices[i]);
+       }
+       
+       
+       sell2
+    }
+}