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.
Pseudocode
the most profitable strategy is to, buy at the lowest price and sell at the highest price.
- loop through array and
- find lowest price
~- find highest price that is after index of lower price~
- find the highest profit
Solution
var maxProfit = function (prices) {
let lowestVal = Infinity;
let maxProfit = 0;
for (let i = 0; i < prices.length; i++) {
if (prices[i] < lowestVal) {
lowestVal = prices[i];
} else if (maxProfit < prices[i] - lowestVal) {
maxProfit = prices[i] - lowestVal;
}
}
return maxProfit;
};
