Problem Description
You are given an array prices where prices[i] is the price of a given stock on the ith day. Find the maximum profit you can achieve. You may complete at most two transactions. Note: You may not engage in multiple transactions simultaneously (i.e., you must sell the stock before you buy again).
Solution
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class Solution {
public int maxProfit(int[] prices) {
//multi-dimentional dp
//state one: n stocks
//state two: buy/sell --> 1/0
//state three: cap --> 0/1/2
int[][][] dp = new int[prices.length][2][3];
for(int i=0; i<dp.length; i++){
for(int j=0; j<dp[0].length; j++){
Arrays.fill(dp[i][j], -1);
}
}
//0 -> starting index
//1 -> allowed to buy (initial state)
//2 -> number of maximum allowed transaction
return helper(prices, 0, 1, 2, dp);
}
public int helper(int[] prices, int idx, int buy, int cap, int[][][] dp){
//if don't have any stock left or if the number of transactions left is 0, return 0
if(idx == prices.length || cap == 0)
return 0;
//if the value of current state: [index, buy/sell, cap] has not been computed previously
if(dp[idx][buy][cap] == -1){
//if we can buy a stock
if(buy == 1){
//we have two options:
//1. buy the current stock and move forward. We reduce the profit by prices[idx] since we will be investing that much.
//2. skip buying, so 0 profit added in this step.
//we will take the maximum profit possible between these two options
dp[idx][buy][cap] = Math.max(
-prices[idx] + helper(prices, idx+1, 0, cap, dp),
0 + helper(prices, idx+1, 1, cap, dp)
);
}else{ //if we cannot buy the current stock (allowed to sell)
//we have two options:
//1. we can sell it with a profit of prices[i]
//2. Skip the current stock and continue to be in same state (not buy)
dp[idx][buy][cap] = Math.max(
prices[idx] + helper(prices, idx+1, 1, cap-1, dp),
0 + helper(prices, idx+1, 0, cap, dp)
);
}
}
return dp[idx][buy][cap];
}
}