Best Time to Buy and Sell Stock II

 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 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.


Using math-


class Solution {
    public int maxProfit(int[] prices) {
        int profit=0;

        for(int i=1;i<prices.length;i++)
        {
            if(prices[i]>prices[i-1])
            {
                profit+=prices[i]-prices[i-1];
            }
        }
        return profit;
    }
}


Using Recursion:

class Solution {
    public int maxProfit(int[] prices) {
        return solve(0, 1, prices); // start at day 0, can buy
    }

    private int solve(int day, int canBuy, int[] prices) {
        // Base case: no more days
        if (day == prices.length) return 0;

        int profit;
        if (canBuy == 1) {
            // Choice: buy or skip
            int buy = -prices[day] + solve(day + 1, 0, prices);
            int skip = solve(day + 1, 1, prices);
            profit = Math.max(buy, skip);
        } else {
            // Choice: sell or skip
            int sell = prices[day] + solve(day + 1, 1, prices);
            int skip = solve(day + 1, 0, prices);
            profit = Math.max(sell, skip);
        }

        return profit;
    }
}


Using DP:-



class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        int[][] dp = new int[n][2];
        for (int[] row : dp) Arrays.fill(row, -1);

        return solve(0, 1, prices, dp);
    }

    private int solve(int day, int canBuy, int[] prices, int[][] dp) {
        if (day == prices.length) return 0;

        if (dp[day][canBuy] != -1) return dp[day][canBuy]; // 1-line memo check

        if (canBuy == 1) {
            dp[day][canBuy] = Math.max(
                -prices[day] + solve(day + 1, 0, prices, dp), // buy
                 solve(day + 1, 1, prices, dp)                // skip
            );
        } else {
            dp[day][canBuy] = Math.max(
                 prices[day] + solve(day + 1, 1, prices, dp), // sell
                 solve(day + 1, 0, prices, dp)                // skip
            );
        }

        return dp[day][canBuy];
    }
}


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