Stock Optimizer!

February 19, 2020

AlgoDaily code challenge, Day 35.

Stock Buy and Sell Optimization

This is a classic technical interview question that I’ve seen a number of times in real life interviews. Let’s say you’re given an array of stock prices, with each element being an integer that represents a price in dollars.

// [ 10, 7, 6, 2, 9, 4 ]

Each index of the array represents a single day, and the the element at that index is the stock price for that given day. This means that the array below represents:

// const prices = [ 10, 7, 6, 2, 9, 4 ];
// Day 0 - a price of `$10`
// Day 1 - `$7`
// Day 2 - `$6` (and so on...)

Given the ability to only buy once and sell once, our goal is to maximize the amount of profit (selling price - purchase price) that we can attain and return that value. Note the only caveat is that that we cannot sell a stock before we buy it, which is important in identifying a solution.

Can you write a stock optimizer method called stockOptimizer? See the following examples for further clarification:

// stockOptimizer([ 10, 7, 6, 2, 9, 4 ])
// 7

For the above example, the max profit is made when we buy on day/index 3(when the price is 2) and sell on day/index 4(when the price is 9) 9 - 2gives us 7, which is returned by the function. Let’s do another example.

stockOptimizer([9, 8, 6, 5, 3])
// 0

From a quick glance, it seems like there’s enough price differences to generate a profit. However, notice that the price drops every day, and thus we can’t have a profit since we’re required to buy before selling.

My Solution:

// Solution:

var stocksArray = [10, 7, 6, 4, 9, 4];
var highestPossibleProfit = 0;

function stockOptimizer(arr) {
  for (i = 0; i < arr.length; i++) {
    for (x = 1; x < arr.length; x++) {
      if (
        arr[i + x] - arr[i] > 0 &&
        arr[i + x] - arr[i] > highestPossibleProfit
      ) {
        highestPossibleProfit = arr[i + x] - arr[i];
      }
    }
  }
}

stockOptimizer(stocksArray);

console.log("Stock Array = " + stocksArray);
console.log("Maximum possible profit = $" + highestPossibleProfit);

Again, pretty simple. Use a nested for loop and keep a “highestPossibleProfit” integer variable. As the for loops run they check if the difference between the two current array elements is > than the current highestPossibleProfit integer.

Yep. Works, simple, clear. What’s not to love?

-mr

Written by Matthew Riddett who lives and works in Victoria BC, building fun and useful things. You can follow him on Twitter