121. Best Time to Buy and Sell Stock using JavaScript

Problem

You are given an array prices where prices[i] is the price of a given stock on the ith day.

You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.

Return the maximum profit you can achieve from this transaction. If you cannot achieve any profit, return 0.

Example 1:

Input: prices = [7,1,5,3,6,4]
Output: 5
Explanation: Buy on day 2 (price = 1) and sell on day 5 (price = 6), profit = 6-1 = 5.
Note that buying on day 2 and selling on day 1 is not allowed because you must buy before you sell.

Example 2:

Input: prices = [7,6,4,3,1]
Output: 0
Explanation: In this case, no transactions are done and the max profit = 0.

Constraints:

• 1 <= prices.length <= 105
• 0 <= prices[i] <= 104

Solution

First attempt : Brute force

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
    let maxprofit = 0;
    for(let i = 0; i < prices.length; i++)
    {
        let currentprofit = 0;
        for(let j = i+1; j<prices.length; j++){
             if(prices[i] >= prices[j]){
                break;
            }
            let currentprofit = (prices[j] - prices[i]);
           // console.log(maxprofit, currentprofit, prices[i], prices[j])
           
            if(currentprofit > maxprofit){
                maxprofit = currentprofit;
                
            }
        }
    }
    return maxprofit;
};

Second attempt : Sliding window method

/**
 * @param {number[]} prices
 * @return {number}
 */
var maxProfit = function(prices) {
    let minPrice = Number.MAX_VALUE;
    let maxProfit = 0;
    
    for(let i = 0; i < prices.length; i++) {
        if(prices[i] < minPrice) {
            minPrice = prices[i];
        } else if(prices[i] - minPrice > maxProfit) {
            maxProfit = prices[i] - minPrice;
        }
    }
    
    return maxProfit;
};