Best Time to Buy and Sell Stock IV

Say you have an array for which the i-th element is the price of a given stock on day i.

Design an algorithm to find the maximum profit. You may complete at most k transactions.

Note: You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again).

Example 1:

Input: [2,4,1], k = 2
Output: 2
Explanation: Buy on day 1 (price = 2) and sell on day 2 (price = 4), profit = 4-2 = 2.

Example 2:

Input: [3,2,6,5,0,3], k = 2
Output: 7
Explanation: Buy on day 2 (price = 2) and sell on day 3 (price = 6), profit = 6-2 = 4.
             Then buy on day 5 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.

class Solution {
    public int maxProfit(int price[], int i, int n, int K, int dp[][][], int ongoing) {
        if (i == n)
            return 0;
        if (K == 0) {
            dp[i][K][ongoing] = 0;
            return 0;
        }
        if (dp[i][K][ongoing] != -1)
            return dp[i][K][ongoing];
        int ans = maxProfit(price, i + 1, n, K, dp, ongoing);
        if (ongoing == 1) {
            int option = maxProfit(price, i + 1, n, K - 1, dp, 0) + price[i];
            ans = Math.max(ans, option);
        } else {
            int option = maxProfit(price, i + 1, n, K, dp, 1) - price[i];
            ans = Math.max(ans, option);
        }
        dp[i][K][ongoing] = ans;
        return ans;
    }

    public int maxProfit(int k, int[] prices) {
        int n = prices.length;
        if (n <= 1)
            return 0;
        // if k >= n/2, then you can make maximum number of transactions.
        // Then this question is similar to infinite transactions
        if (k >= n / 2) {
            int T_ik0 = 0, T_ik1 = Integer.MIN_VALUE;
    
            for (int price : prices) {
                int T_ik0_old = T_ik0;
                // If we want 0 stocks in our hand at end of ith day
                // Either take ans from 0 stock i-1th day
                // Or take ans from 1 stock i-1th day and sell it
                T_ik0 = Math.max(T_ik0, T_ik1 + price);
                // If we want 1 stock in our hand at end of ith day
                // Either take ans from 1 stock i-1th day
                // Or take ans from 0 stock i-1th day and but 1 stock at ith day
                T_ik1 = Math.max(T_ik1, T_ik0_old - price);
            }
    
            return T_ik0;
        }
        int dp[][][] = new int[n + 1][k + 1][2];
        for (int i = 0; i <= n; i++) {
            for (int j = 0; j <= k; j++) {
                dp[i][j][0] = -1;
                dp[i][j][1] = -1;
            }
        }
        return maxProfit(prices, 0, n, k, dp, 0);
    }
}