leetcode 16: Best Time to Buy and Sell Stock III

Say you have an array for which the i^th 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).

key points: 1. divide to two parts. get the sum of two parts. each part can use the same approach from Sell Stock I.

time complexity    O(n^2).

Problems. can not pass large test set cause of exceeding time limit.    (maybe more efficient algorithm exists.)

class Solution {
public:
    int partialMax(vector<int> &prices, int start, int end) {
        if( start >= end) return 0;
        
        int maxPrice = 0;
        int minIndex = start;
        
        for( int i=start+1; i<=end; i++) {
            if( prices[i] - prices[minIndex] > maxPrice) {
                maxPrice = prices[i] - prices[minIndex];
            }   
            
            if( prices[i] < prices[minIndex]) {
                minIndex =  i;
            }
        }
        
        return maxPrice;
        
    }

    int maxProfit(vector<int> &prices) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int sz = prices.size();
        if( sz < 2 ) return 0;
        if(sz<3) {
            return prices[1] - prices[0] > 0 ? prices[1]-prices[0] : 0;
        }
        
        int max=INT_MIN;
        
        for(int i=1; i<sz-1; i++) {
            int temp = partialMax(prices, 0, i) + partialMax(prices, i, sz-1);
            if(temp > max) max = temp;
        }
        
        return max;
    }
};

public class Solution {
    public int maxProfit(int[] prices) {
        // Start typing your Java solution below
        // DO NOT write main() function
        int sz = prices.length;
        if(sz<=1) return 0;
        
        int[] dp1 = new int[sz];
        int[] dp2 = new int[sz];
        
        int min = prices[0];
        int max = prices[sz-1];
        int profit = 0;
        
        for(int i=1; i<sz; i++) {
            dp1[i] = Math.max( profit, prices[i]-min);
            min = Math.min( prices[i], min);
        }
        
        profit = 0;
        for( int i=sz-2; i>=0; i--) {
            dp2[i] = Math.max( profit, max - prices[i] );
            max = Math.max( prices[i], max);
        }
        
        int rel=0;
        for( int i=0; i<sz; i++) {
            rel = Math.max(rel, dp1[i]+dp2[i]);
        }
        return rel;
    }
};