Best Time to Buy and Sell Stock IV

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

Find the maximum profit you can achieve. You may complete at most k transactions: i.e. you may buy at most k times and sell at most k times.

Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).

Example 1:

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

Example 2:

Input: k = 2, prices = [3,2,6,5,0,3]
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.

Constraints:

• 1 <= k <= 100
• 1 <= prices.length <= 1000
• 0 <= prices[i] <= 1000

Approach

• if k >= n / 2 then we know we can do unlimited transaction so we can do the simple profit calculation did for III
• for else scenario we need to maintain buy and sell array
• O(n*k), O(k)

class Solution {
    public int maxProfit(int k, int[] prices) {
        int n = prices.length;
 
        if (k >= n/2) {
            int profit = 0;
            for (int i = 1; i < n; i++) {
                if (prices[i] > prices[i-1])
                    profit += prices[i] - prices[i-1];
            }
 
            return profit;
        }
 
        int[] buy = new int[k+1];
        int[] sell = new int[k+1];
        Arrays.fill(buy, Integer.MIN_VALUE);
 
        for (int p: prices) {
            for (int i = 1; i <= k; i++) {
                buy[i] = Math.max(buy[i], sell[i-1] - p);
                sell[i] = Math.max(sell[i], buy[i] + p);
            }
        }
 
        return sell[k];
    }
}