Best Time to Buy and Sell Stock

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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.

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;
};

Time and Space Complexity

Time

• Loop through the array once O(N)

• Total - O(N)

Space

• Storing lowest price and max profit O(1)

• Total - O(1)