package com.fishercoder.solutions;

/**
 * 188. Best Time to Buy and Sell Stock IV
 *
 * Say you have an array for which the ith 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).
 */
public class _188 {

    /**credit: https://discuss.leetcode.com/topic/8984/a-concise-dp-solution-in-java*/
    public int maxProfit(int k, int[] prices) {
        int len = prices.length;
        if (k >= len / 2) {
            return quickSolve(prices);
        }

        int[][] t = new int[k + 1][len];
        for (int i = 1; i <= k; i++) {
            int tmpMax =  -prices[0];
            for (int j = 1; j < len; j++) {
                t[i][j] = Math.max(t[i][j - 1], prices[j] + tmpMax);
                tmpMax =  Math.max(tmpMax, t[i - 1][j - 1] - prices[j]);
            }
        }
        return t[k][len - 1];
    }


    private int quickSolve(int[] prices) {
        int len = prices.length;
        int profit = 0;
        for (int i = 1; i < len; i++) {
            // as long as there is a price gap, we gain a profit.
            if (prices[i] > prices[i - 1]) {
                profit += prices[i] - prices[i - 1];
            }
        }
        return profit;
    }

}

package com.fishercoder.solutions;

/**

 188. Best Time to Buy and Sell Stock IV

 Say you have an array for which the ith 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.

 */
public class _188 {
  public static class Solution1 {
    /** credit: https://discuss.leetcode.com/topic/8984/a-concise-dp-solution-in-java */
    public int maxProfit(int k, int[] prices) {
      int len = prices.length;
      if (k >= len / 2) {
        return quickSolve(prices);
      }

      int[][] t = new int[k + 1][len];
      for (int i = 1; i <= k; i++) {
        int tmpMax = -prices[0];
        for (int j = 1; j < len; j++) {
          t[i][j] = Math.max(t[i][j - 1], prices[j] + tmpMax);
          tmpMax = Math.max(tmpMax, t[i - 1][j - 1] - prices[j]);
        }
      }
      return t[k][len - 1];
    }

    private int quickSolve(int[] prices) {
      int len = prices.length;
      int profit = 0;
      for (int i = 1; i < len; i++) {
        // as long as there is a price gap, we gain a profit.
        if (prices[i] > prices[i - 1]) {
          profit += prices[i] - prices[i - 1];
        }
      }
      return profit;
    }
  }
}

refactor 188