JavaScript Program to Find Maximum Profit by Buying and Selling a Share at Most Twice using Array

function maxProfit(price, n) {

    // Create profit array and
    // initialize it as 0
    let profit = new Array(n);
    for (let i = 0; i < n; i++)
        profit[i] = 0;

    /* Get the maximum profit with
    only one transaction
    allowed. After this loop,
    profit[i] contains maximum
    profit from price[i..n-1]
    using at most one trans. */
    let max_price = price[n - 1];
    for (let i = n - 2; i >= 0; i--) {

        // Max_price has maximum
        // of price[i..n-1]
        if (price[i] > max_price)
            max_price = price[i];

        // We can get profit[i] by taking maximum of:
        // a) previous maximum, i.e., profit[i+1]
        // b) profit by buying at price[i] and selling at
        // max_price
        profit[i] = Math.max(profit[i + 1],
            max_price - price[i]);
    }

    // Get the maximum profit with
    // two transactions allowed
    // After this loop, profit[n-1]
    // contains the result
    let min_price = price[0];
    for (let i = 1; i < n; i++) {

        // min_price is minimum price
        // in price[0..i]
        if (price[i] < min_price)
            min_price = price[i];

        // Maximum profit is maximum of:
        // a) previous maximum, i.e., profit[i-1]
        // b) (Buy, Sell) at (min_price, price[i]) and add
        // profit of other trans. stored in profit[i]
        profit[i] = Math.max(profit[i - 1],
            profit[i] + (price[i] - min_price));
    }
    let result = profit[n - 1];

    return result;
}

// Driver code
let price = [2, 30, 15, 10, 8, 25, 80];
let n = price.length;

console.log("Maximum Profit = " +
    maxProfit(price, n));

function f(idx, buy, prices, cap, n) {

    if (cap == 0) {
        return 0;
    }

    if (idx == n) {
        return 0;
    }

    let profit = 0;

    // You can either buy or not buy
    if (buy == 0) {
        profit = Math.max(
            -prices[idx] + f(idx + 1, 1, prices, cap, n),
            f(idx + 1, 0, prices, cap, n)
        );
    }
    // You can either sell or not sell
    else {
        profit = Math.max(
            prices[idx] + f(idx + 1, 0, prices, cap - 1, n),
            f(idx + 1, 1, prices, cap, n)
        );
    }

    return profit;
}

function maxtwobuysell(prices, n) {
    return f(0, 0, prices, 2, n);
}

const arr = [2, 30, 15, 10, 8, 25, 80];
const size = arr.length;
console.log(maxtwobuysell(arr, size));