Best Time to Buy and Sell Stock III

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:
At first, I tried to use a similar way as Best Time to Buy and Sell Stock I. I used divide and conquer, divide the array into two parts: [0...i][i...n-1], but the recursion process is time-consuming, I got time limit exceed in leetcode online judge.
Think one more step, we use two arrays: l[i] donates max profit for [0...i] and r[i] donates max profit for [i...n-1]. In order to get r[i], we are going to iterate from right to left and record max sell price, since we must sell stock before buy.

public class Solution {
    public int maxProfit(int[] prices) {
        int n = prices.length;
        if(n == 0) return 0;
        int[] l = new int[n];
        int[] r = new int[n];
        
        // max profit from 0 to i;
        l[0] = 0;
        int minBuy = prices[0];
        for(int i = 1; i < n; i++){
            // update max profit
            l[i] = Math.max(l[i-1], prices[i] - minBuy);
            minBuy = Math.min(minBuy, prices[i]);
        }
        
        // max profit from i to n;
        r[n-1] = 0;
        int maxSell = prices[n-1];
        for(int i = n-2; i >= 0; i--){
            // update max profit
            r[i] = Math.max(r[i+1], maxSell - prices[i]);
            maxSell = Math.max(maxSell, prices[i]);
        }
        
        int result = 0;
        for(int i = 0; i < n; i++){
            result = Math.max(result, l[i] + r[i]);
        }
        return result;
    }
}