Skip to main content

Buy and Sell Stock - II

Problem Statement:​

You are given an integer array prices where prices[i] is the price of a given stock on the ith day.

On each day, you may decide to buy and/or sell the stock. You can only hold at most one share of the stock at any time. However, you can buy it then immediately sell it on the same day.

Find and return the maximum profit you can achieve.

  • Example:

    Example 1:

    Input: prices = [7,1,5,3,6,4]
    Output: 7
    Explanation: Buy on day 2 (price = 1) and sell on day 3 (price = 5), profit = 5-1 = 4.
    Then buy on day 4 (price = 3) and sell on day 5 (price = 6), profit = 6-3 = 3.
    Total profit is 4 + 3 = 7.
    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.
    Total profit is 4.
    Example 3:

    Input: prices = [7,6,4,3,1]
    Output: 0
    Explanation: There is no way to make a positive profit, so we never buy the stock to achieve the maximum profit of 0

✅ Solution: Memoization​

class Solution {
public:
    int solve(int ind, vector<int> prices, int buyOrSell, vector<vector<int>> &dp){
        if(ind == prices.size())
            return 0;

        if(dp[ind][buyOrSell] != -1) return dp[ind][buyOrSell];

        int op1 = 0, op2 = 0;
        if(buyOrSell == 0){
            // Either skip buying or buy at current price
            op1 = solve(ind + 1, prices, 0, dp);
            op2 = -prices[ind] + solve(ind + 1, prices, 1, dp);
        }
        if(buyOrSell == 1){
            // Either skip selling or sell at current price
            op1 = solve(ind + 1, prices, 1, dp);
            op2 = prices[ind] + solve(ind + 1, prices, 0, dp);
        }

        return dp[ind][buyOrSell] = max(op1, op2);
    }

    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp(n + 1, vector<int>(2, -1));
        return solve(0, prices, 0, dp); // 0 = buy allowed
    }
};

✅ Solution: Tabulation​

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<vector<int>> dp(n + 1, vector<int>(2, 0)); // dp[i][0]: buy, dp[i][1]: sell

        for(int i = n - 1; i >= 0; i--){
            for(int buy = 0; buy < 2; buy++){
                if(buy == 0){
                    dp[i][buy] = max(dp[i + 1][0], -prices[i] + dp[i + 1][1]);
                } else {
                    dp[i][buy] = max(dp[i + 1][1], prices[i] + dp[i + 1][0]);
                }
            }
        }

        return dp[0][0];
    }
};

✅ Solution: Space Optimized​

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        int n = prices.size();
        vector<int> next(2, 0), curr(2, 0);

        for(int i = n - 1; i >= 0; i--){
            for(int buy = 0; buy < 2; buy++){
                if(buy == 0){
                    curr[buy] = max(next[0], -prices[i] + next[1]);
                } else {
                    curr[buy] = max(next[1], prices[i] + next[0]);
                }
            }
            next = curr;
        }

        return curr[0];
    }
};

📝 How It Works​

  • You can buy and sell stocks multiple times, but only one stock at a time.
  • For each day and state (buy/sell), decide:
    • If buy allowed: buy or skip.
    • If sell allowed: sell or skip.
  • Use dp[ind][buyOrSell] to memoize answers to subproblems.

🧩 Key Formula / Recurrence​

Let dp[i][buy] represent max profit on day i with buy permission:

if(buy == 0)
    dp[i][0] = max(dp[i+1][0], -prices[i] + dp[i+1][1])
else
    dp[i][1] = max(dp[i+1][1], prices[i] + dp[i+1][0])

⏱️ Time & Space Complexity​

ApproachTimeSpace
MemoizationO(n × 2)O(n × 2)
TabulationO(n × 2)O(n × 2)
Space OptimizedO(n × 2)O(1 × 2) = O(1) ✅

⚠️ Edge Cases​

  • Empty array → return 0
  • All decreasing prices → return 0
  • All increasing → buy on day 0, sell on last day (sum of differences)

💡 Other Approaches​

MethodTimeSpaceNotes
Greedy (sum of all increasing differences)O(n)O(1)Efficient for this problem

🔁 Related Problems​

💬

Discussion & Doubts