LeetCode Question of the Day (309. Best Time to Buy and Sell Stock with Cooldown)

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 as many transactions as you like (i.e., buy one and sell one share of the stock multiple times) with the following restrictions:

After you sell your stock, you cannot buy stock on the next day (i.e., cooldown one day).
Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

Input: prices = [1,2,3,0,2]
Output: 3

Explanation: transactions = [buy, sell, cooldown, buy, sell]

Example 2:

Input: prices = [1]
Output: 0

Constraints:

• 1 <= prices.length <= 5000
• 0 <= prices[i] <= 1000

---

当 prices[i] >= prices[i-1]的时候, 我们有两个选择, One is to sell the current stock, 然后跳到 i+2(because after the sale 1 天的冷静期), The other is not for sale, then walk normally i+1. We maintain both the current maximum and minimum prices, Similar to a monotonically increasing queue, Used to record the current best buying time and best selling time

---

use std::collections::HashMap;

impl Solution {
    
    fn dp(
        prices: &Vec<i32>,
        i: usize,
        min: i32,
        max: i32,
        cache: &mut HashMap<(usize, i32, i32), i32>,
    ) -> i32 {
    
        if i >= prices.len() {
    
            return max - min;
        }
        let curr = prices[i];
        if min == -1 && max == -1 {
    
            let ans = if let Some(c) = cache.get(&(i + 1, curr, curr)) {
    
                *c
            } else {
    
                Solution::dp(prices, i + 1, curr, curr, cache)
            };
            cache.insert((i, -1, -1), ans);
            return ans;
        }
        if curr >= max {
    
            let sell = if let Some(c) = cache.get(&(i + 2, -1, -1)) {
    
                *c
            } else {
    
                Solution::dp(prices, i + 2, -1, -1, cache)
            } + curr
                - min;
            let keep = if let Some(c) = cache.get(&(i + 1, min, max)) {
    
                *c
            } else {
    
                Solution::dp(prices, i + 1, min, max, cache)
            };
            let ans = sell.max(keep);
            cache.insert((i, min, max), ans);
            return ans;
        }
        if curr <= min {
    
            let ans = if let Some(c) = cache.get(&(i + 1, curr, curr)) {
    
                *c
            } else {
    
                Solution::dp(prices, i + 1, curr, curr, cache)
            };
            cache.insert((i, min, max), ans);
            return ans;
        } else {
    
            let ans = if let Some(c) = cache.get(&(i + 1, min, curr)) {
    
                *c
            } else {
    
                Solution::dp(prices, i + 1, min, curr, cache)
            };
            cache.insert((i, min, max), ans);
            return ans;
        }
    }
    pub fn max_profit(prices: Vec<i32>) -> i32 {
    
        Solution::dp(&prices, 0, -1, -1, &mut HashMap::new())
    }
}