Best Time to Buy and Sell Stock III

Problem statement

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).

Example 1:

Input: prices = [3,3,5,0,0,3,1,4]Output: 6Explanation: Buy on day 4 (price = 0) and sell on day 6 (price = 3), profit = 3-0 = 3.Then buy on day 7 (price = 1) and sell on day 8 (price = 4), profit = 4-1 = 3.

Example 2:

Input: prices = [1,2,3,4,5]Output: 4Explanation: Buy on day 1 (price = 1) and sell on day 5 (price = 5), profit = 5-1 = 4.Note that you cannot buy on day 1, buy on day 2 and sell them later, as you are engaging multiple transactions at the same time. You must sell before buying again.

Example 3:

Input: prices = [7,6,4,3,1]Output: 0Explanation: In this case, no transaction is done, i.e. max profit = 0.

Constraints:

• 1 <= prices.length <= 105
• 0 <= prices[i] <= 105

My solution

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
    const leftRightProfit = [0];
    let minPrice = prices[0]
    
    for (let i = 1; i < prices.length; i++) {
        const price = prices[i]
        minPrice = Math.min(minPrice, price);
        leftRightProfit.push(
            Math.max(leftRightProfit[i - 1], price - minPrice)
        )
    }
    
    const rightLeftProfit = [0]
    let maxPrice = prices[prices.length - 1]
    
    for (let i = prices.length - 2; i >= 0; i--) {
        const price = prices[i]
        maxPrice = Math.max(maxPrice, price);
        rightLeftProfit.unshift(
            Math.max(rightLeftProfit[0], maxPrice - price)
        )
    }
    
    let profit = 0;
    for (let i = 0; i < prices.length; i++) {
        profit = Math.max(profit, leftRightProfit[i] + rightLeftProfit[i])
    }
    
    return profit;
};