bestTimeToBuyAndSellStock.III.cpp

// Source : https://oj.leetcode.com/problems/best-time-to-buy-and-sell-stock-iii/
// Author : Hao Chen
// Date   : 2014-08-22

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


class Solution {
public:
    // Dynamic Programming
    //
    //     Considering prices[n], and we have a position "i", we could have
    //       1) the maxProfit1 for prices[0..i]  
    //       2) the maxProfit2 for proices[i..n]
    //
    //    So, 
    //      for 1) we can go through the prices[n] forwardly.
    //          forward[i] = max( forward[i-1], price[i] - lowestPrice[0..i] ) 
    //      for 2) we can go through the prices[n] backwoardly.
    //          backward[i] = max( backward[i+1], highestPrice[i..n] - price[i]) 
    //
    int maxProfit(vector<int> &prices) {
        if (prices.size()<=1) return 0;
        
        int n = prices.size();
        
        vector<int> forward(n);
        forward[0] = 0;
        int lowestBuyInPrice = prices[0];
        for(int i=1; i<n; i++){
            forward[i] = max(forward[i-1], prices[i] - lowestBuyInPrice);
            lowestBuyInPrice = min(lowestBuyInPrice, prices[i]);
        }
        
        vector<int> backward(n);
        backward[n-1] = 0;
        int highestSellOutPrice = prices[n-1];
        for(int i=n-2; i>=0; i--){
            backward[i] = max(backward[i+1], highestSellOutPrice - prices[i]);
            highestSellOutPrice = max(highestSellOutPrice, prices[i]);
        }
        
        int max_profit = 0;
        for(int i=0; i<n; i++){
            max_profit = max(max_profit, forward[i]+backward[i]);
        }
        
        return max_profit;
    }
};