Best Time to Buy and Sell Stock III (Java)

LeetCode

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).

Solution: Divide and Conquer, DP

O(n^2) solution is easy came out, we can use O(n) time to get max profit depend on the solution of Best Time to Buy and Sell Stock I. so we can just dived the whole prices array at every point , try to calculate current max profit from left and right and then add them together is what we want. However, there are many repeat calculation when calculate left max profit and right max profit. So we can apply DP to record max profits for each left side and right side. then add them together at each point use one for loop

	/*
	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).
	*/
	public class BestTimetoBuyandSellStockIII {
	public int maxProfit(int[] prices) {
	if (prices==null || prices.length==0){
	return 0;
	}
	// record first max profit at first buy and sell
	int[] leftProfits=new int[prices.length];
	// record second max profit at second buy and sell
	int[] rightProfits=new int[prices.length];
	int max=0;
	buildProfitsArray(prices, leftProfits, rightProfits);
	// max profit made by left side + max profit made by rigth side is the max profit made by two buy and sell
	for (int i=0; i<prices.length; i++){
	max=Math.max(max, leftProfits[i]+rightProfits[i]);
	}
	return max;
	}
	private void buildProfitsArray(int [] prices, int[] leftProfits,int[] rightProfits){
	int min=Integer.MAX_VALUE;
	// max profits could make by buy stock at i from 0<=i<n and sell at i
	for (int i=0; i<leftProfits.length; i++){
	if (prices[i]<min){
	min=prices[i];
	}
	leftProfits[i]=i==0?0:Math.max(leftProfits[i-1], prices[i]-min);
	}
	int max=Integer.MIN_VALUE;
	// max profits could make by buy stock at i and sell at j, i<=j<n
	for (int i=rightProfits.length-1; i>=0; i--){
	if (prices[i]>max){
	max=prices[i];
	}
	rightProfits[i]=i==rightProfits.length-1?0: Math.max(rightProfits[i+1], max-prices[i]);
	}
	}
	}

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