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update 714 java
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‎doc/cheatsheet/dp.md‎

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@@ -11,6 +11,8 @@ problems by combining the solutions to subproblems
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- https://predoc.dlc.ntu.edu.tw/viewer?embedded=true&url=https%3A%2F%2Fcool.ntu.edu.tw%2Fcourses%2F8583%2Ffiles%2F1165602%2Fdownload%3Fverifier%3DnlJ3s1a9TTmgYQzbJgj9vnrGlKKZB4w0wUZyEKgm
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### 0-1) Types
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- 2 DP state:
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- LC 714
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### 0-2) Pattern
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@@ -185,4 +187,98 @@ private int findMin(int a, int b, int c){
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return c;
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}
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}
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```
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### 2-3) Best Time to Buy and Sell Stock with Transaction Fee
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```java
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// java
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// LC 714
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// V0-1
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// IDEA: DP (gpt)
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/**
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* Solution Explanation:
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*
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*
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* - Use two variables to represent the state:
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* 1. hold: The maximum profit achievable
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* while holding a stock at day i.
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*
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* 2. cash: The maximum profit achievable
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* while not holding a stock at day i.
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*
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* - Transition equations:
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* - If holding a stock:
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* hold = max(hold, cash - price[i])
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*
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* NOTE: 2 cases we hold th stock: 1) already hold from previous day 2) buy a new stock today
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* (`hold`: You already held the stock from a previous day -> If you decided not to make any changes today, then the profit remains the same as the previous hold.)
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* (`cash - price[i]`: You buy the stock today -> To buy the stock today, you need to spend money, reducing your profit. The cost to buy the stock is prices[i]. However, the amount of money you can spend is the maximum profit you had when you were not holding a stock previously (cash).)
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*
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* (You either keep holding or buy a new stock.)
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* - If not holding a stock:
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* cash = max(cash, hold + price[i] - fee)
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*
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*
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* (You either keep not holding or sell the stock and pay the fee.)
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* - Initialize:
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* - hold = -prices[0] (If you buy the stock on the first day).
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* - cash = 0 (You haven’t made any transactions yet).
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*
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*/
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/**
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* Example Walkthrough:
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*
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* Input:
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* • Prices: [1, 3, 2, 8, 4, 9]
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* • Fee: 2
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*
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* Steps:
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* 1. Day 0:
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* • hold = -1 (Buy the stock at price 1).
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* • cash = 0.
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* 2. Day 1:
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* • cash = max(0, -1 + 3 - 2) = 0 (No selling since profit is 0).
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* • hold = max(-1, 0 - 3) = -1 (No buying since it’s already better to hold).
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* 3. Day 2:
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* • cash = max(0, -1 + 2 - 2) = 0.
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* • hold = max(-1, 0 - 2) = -1.
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* 4. Day 3:
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* • cash = max(0, -1 + 8 - 2) = 5 (Sell at price 8).
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* • hold = max(-1, 5 - 8) = -1.
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* 5. Day 4:
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* • cash = max(5, -1 + 4 - 2) = 5.
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* • hold = max(-1, 5 - 4) = 1.
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* 6. Day 5:
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* • cash = max(5, 1 + 9 - 2) = 8 (Sell at price 9).
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* • hold = max(1, 5 - 9) = 1.
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*
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* Output:
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* • cash = 8 (Max profit).
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*
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*/
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public int maxProfit_0_1(int[] prices, int fee) {
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// Edge case
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if (prices == null || prices.length == 0) {
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return 0;
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}
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// Initialize states
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int hold = -prices[0]; // Maximum profit when holding a stock
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int cash = 0; // Maximum profit when not holding a stock
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// Iterate through prices
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for (int i = 1; i < prices.length; i++) {
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/**
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* NOTE !!! there are 2 dp equations (e.g. cash, hold)
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*/
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// Update cash and hold states
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cash = Math.max(cash, hold + prices[i] - fee); // Sell the stock
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hold = Math.max(hold, cash - prices[i]); // Buy the stock
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}
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// The maximum profit at the end is when not holding any stock
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return cash;
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}
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```

‎leetcode_java/src/main/java/LeetCodeJava/DynamicProgramming/BestTimeToBuyAndSellStockWithTransactionFee.java‎

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*/
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public class BestTimeToBuyAndSellStockWithTransactionFee {
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// V0
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// public int maxProfit(int[] prices, int fee) {
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//
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// }
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// V0
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// public int maxProfit(int[] prices, int fee) {
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//
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// }
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// V0-1
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// IDEA: DP (gpt)
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/**
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* Solution Explanation:
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*
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*
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* - Use two variables to represent the state:
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* 1. hold: The maximum profit achievable
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* while holding a stock at day i.
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*
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* 2. cash: The maximum profit achievable
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* while not holding a stock at day i.
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*
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* - Transition equations:
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* - If holding a stock:
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* hold = max(hold, cash - price[i])
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*
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* NOTE: 2 cases we hold th stock: 1) already hold from previous day 2) buy a new stock today
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* (`hold`: You already held the stock from a previous day -> If you decided not to make any changes today, then the profit remains the same as the previous hold.)
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* (`cash - price[i]`: You buy the stock today -> To buy the stock today, you need to spend money, reducing your profit. The cost to buy the stock is prices[i]. However, the amount of money you can spend is the maximum profit you had when you were not holding a stock previously (cash).)
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*
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* (You either keep holding or buy a new stock.)
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* - If not holding a stock:
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* cash = max(cash, hold + price[i] - fee)
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*
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*
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* (You either keep not holding or sell the stock and pay the fee.)
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* - Initialize:
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* - hold = -prices[0] (If you buy the stock on the first day).
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* - cash = 0 (You haven’t made any transactions yet).
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*
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*/
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/**
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* Example Walkthrough:
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*
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* Input:
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* • Prices: [1, 3, 2, 8, 4, 9]
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* • Fee: 2
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*
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* Steps:
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* 1. Day 0:
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* • hold = -1 (Buy the stock at price 1).
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* • cash = 0.
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* 2. Day 1:
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* • cash = max(0, -1 + 3 - 2) = 0 (No selling since profit is 0).
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* • hold = max(-1, 0 - 3) = -1 (No buying since it’s already better to hold).
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* 3. Day 2:
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* • cash = max(0, -1 + 2 - 2) = 0.
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* • hold = max(-1, 0 - 2) = -1.
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* 4. Day 3:
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* • cash = max(0, -1 + 8 - 2) = 5 (Sell at price 8).
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* • hold = max(-1, 5 - 8) = -1.
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* 5. Day 4:
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* • cash = max(5, -1 + 4 - 2) = 5.
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* • hold = max(-1, 5 - 4) = 1.
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* 6. Day 5:
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* • cash = max(5, 1 + 9 - 2) = 8 (Sell at price 9).
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* • hold = max(1, 5 - 9) = 1.
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*
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* Output:
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* • cash = 8 (Max profit).
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*
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*/
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public int maxProfit_0_1(int[] prices, int fee) {
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// Edge case
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if (prices == null || prices.length == 0) {
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return 0;
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}
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// Initialize states
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int hold = -prices[0]; // Maximum profit when holding a stock
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int cash = 0; // Maximum profit when not holding a stock
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// Iterate through prices
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for (int i = 1; i < prices.length; i++) {
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/**
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* NOTE !!! there are 2 dp equations (e.g. cash, hold)
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*/
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// Update cash and hold states
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cash = Math.max(cash, hold + prices[i] - fee); // Sell the stock
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hold = Math.max(hold, cash - prices[i]); // Buy the stock
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}
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// The maximum profit at the end is when not holding any stock
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return cash;
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}
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// V1
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// https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/solutions/3667440/beats-100-c-java-python-beginner-friendl-rpgh/

‎leetcode_java/src/main/java/dev/workspace6.java‎

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return null;
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}
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// LC 714
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// 7.17 pm - 7.30 pm
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// https://leetcode.com/problems/best-time-to-buy-and-sell-stock-with-transaction-fee/description/
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/**
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*
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*
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* IDEA: DP
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*
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*
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* 1. transaction fee need to be paid at EVERY sale
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* 2.
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*
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*
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* Exp 1)
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*
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* Input: prices = [1,3,2,8,4,9], fee = 2
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* -> Output: 8
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*
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*
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* [1,3,2,8,4,9], fee = 2
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* b s (rev = (3-1-2) = 0)
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* b s (rev = (8-1-2) = 5
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*
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*/
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public int maxProfit(int[] prices, int fee) {
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int res = 0;
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// edge
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if (prices.length == 1){
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return 0;
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}
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if(prices.length == 2){
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if(prices[1] > prices[0]){
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return Math.max(0, prices[1] - prices[0] - fee);
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}
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}
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// dp
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int[] dp = new int[prices.length+1];
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// fill dp with 0
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Arrays.fill(dp, 0); // ??
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// init state
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dp[0] = 0;
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if(prices[1] > prices[0]){
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dp[1] = Math.max(0, prices[1] - prices[0] - fee);
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}
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// 2 option:
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// if not buy: buy or not buy
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// if already buy: sale or hold
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return res;
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}
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}

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