Problem 309: Best Time to Buy and Sell Stock with Cooldown

Given an integer array prices where prices[i] represents the stock price on day i. Design an algorithm to calculate the maximum profit under the constraint that you cannot buy stock on the day after selling (one-day cooldown period).

Dynamic Programming Approach

The key to solving this problem lies in properly defining states. Building upon the two-state model from the unlimited transactions problem, we introduce an additional cooldown state.

State Definition

Define dp[i][j] as the maximum cash on day i under state j:

State Transitions

State 0 (Holding):

• Continue holding: dp[i-1][0]
• Buy today from cooldown: dp[i-1][3] - prices[i]
• Buy today from regular selling state: dp[i-1][1] - prices[i]

State 1 (Regular No-Hold):

• Stay in this state: dp[i-1][1]
• Transition from cooldown: dp[i-1][3]

State 2 (Sold Today):

• Must have been holding yesterday: dp[i-1][0] + prices[i]

State 3 (Cooldown):

• Must have sold yesterday: dp[i-1][2]

Implementation

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int days = prices.size();
        if (days == 0) return 0;
        
        vector<array<int, 4>> dp(days);
        dp[0][0] = -prices[0];  // Buy on day 0
        dp[0][1] = dp[0][2] = dp[0][3] = 0;
        
        for (int i = 1; i < days; i++) {
            dp[i][0] = max({dp[i-1][0], dp[i-1][3] - prices[i], dp[i-1][1] - prices[i]});
            dp[i][1] = max(dp[i-1][1], dp[i-1][3]);
            dp[i][2] = dp[i-1][0] + prices[i];
            dp[i][3] = dp[i-1][2];
        }
        
        return max({dp[days-1][1], dp[days-1][2], dp[days-1][3]});
    }
};

• Time Complexity: O(n)
• Space Complexity: O(n)

Problem 714: Best Time to Buy and Sell Stock with Transaction Fee

Given an array prices where prices[i] is the stock price on day i, and an integer fee representing the transaction cost. You may complete unlimited transactions, but each transaction incurs a one-time fee. You must sell before buying again.

Dynamic Programming Approach

This problem extends the unlimited transactions model by incorporating a transaction cost. The state space remains minimal with just two states.

State Definition

State Transitions

Holding State:

• Keep holding: dp[i-1][0]
• Buy today: dp[i-1][1] - prices[i]

Not-Holding State:

• Keep not holding: dp[i-1][1]
• Sell today: dp[i-1][0] + prices[i] - fee

Implementation

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int days = prices.size();
        vector<array<int, 2>> dp(days);
        
        dp[0][0] = -prices[0];  // Initial purchase
        dp[0][1] = 0;
        
        for (int i = 1; i < days; i++) {
            dp[i][0] = max(dp[i-1][0], dp[i-1][1] - prices[i]);
            dp[i][1] = max(dp[i-1][1], dp[i-1][0] + prices[i] - fee);
        }
        
        return max(dp[days-1][0], dp[days-1][1]);
    }
};

• Time Complexity: O(n)
• Space Complexity: O(n)

Space Optimization

Since each state only depends on the previous day, we can reduce space to O(1):

class Solution {
public:
    int maxProfit(vector<int>& prices, int fee) {
        int hold = -prices[0];
        int cash = 0;
        
        for (int i = 1; i < prices.size(); i++) {
            int prevHold = hold;
            hold = max(hold, cash - prices[i]);
            cash = max(cash, prevHold + prices[i] - fee);
        }
        
        return cash;
    }
};