188. Best Time to Buy and Sell Stock IV.md

188. Best Time to Buy and Sell Stock IV

Question

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.

Solution

• Method 1: dp
  1. We need to pay attention to the initialization.
  2. Transaction function is same as 123. Best Time to Buy and Sell Stock III.
  3. If k is larger than prices length, we treat it as question 122. Best Time to Buy and Sell Stock II.

   class Solution {
   	public int maxProfit(int k, int[] prices) {
   		if(k == 0 || prices == null || prices.length <= 1) return 0;
   		int len = prices.length;
   		if(k > len) return maxProfit(prices);
   		int[][] buys = new int[len + 1][k + 1];
   		int[][] sells = new int[len + 1][k + 1];
   		for(int i = 2; i <= k; i++){
   			for(int j = 0; j < i; j++){
   				buys[j][i] = Integer.MIN_VALUE;
   			}
   		}
   		int sum = 0;
   		for(int i = 1; i <= k; i++){
   			sum -= prices[i - 1];
   			buys[i][i] = sum;
   		}
   		for(int i = 2; i <= len; i++){
   			for(int j = 1; j <= k; j++){
   				buys[i][j] = Math.max(buys[i - 1][j], sells[i - 1][j - 1] - prices[i - 1]);
   				sells[i][j] = Math.max(sells[i - 1][j], buys[i - 1][j] + prices[i - 1]);
   			}
   		}        
   		int profit = 0;
   		for(int i = 0; i <= k; i++){
   			profit = Math.max(profit, sells[len][i]);
   		}
   		return profit;
   	}
   	
   	private int maxProfit(int[] prices) {
   			int len = prices.length;
   			int[] buys = new int[len + 1];
   			int[] sells = new int[len + 1];
   			buys[1] = -prices[0];
   			for(int i = 2; i <= len; i++){
   				buys[i] = Math.max(buys[i - 1], sells[i - 1] - prices[i - 1]);
   				sells[i] = Math.max(sells[i - 1], buys[i - 1] + prices[i - 1]);
   			}
   			return sells[len];
   		}
   }