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    • Stock maximum profit
    The question is as follows:
    You will be given a list of stock prices for a given day and your goal is to return the maximum profit that could have been made by buying a stock at the given price and then selling the stock later on. For example if the input is: [45, 24, 35, 31, 40, 38, 11] then your program should return 16 because if you bought the stock at $24 and sold it at $40, a profit of $16 was made and this is the largest profit that could be made. If no profit could have been made, return -1.



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    mrdaniel published this on 1/3/16 | javascript, bootcamp, Fullstack Academy, Hack Reactor, Google
    Comments
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    • 4
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    • I am confused as to why the "continue" keyword is used here. It seems to me that the code would (and does) execute just fine without it? Was there a specific reason it was left in there? Thanks so much!
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    • 1
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    • This is how i did it. var profitArray is map of the agumented array. It takes every value a turns it into the profit one would make if he bought the stocks that day and sold them on the most expensive day left. profitArray.sort(function(a, b){return b-a})[0] simply returns the highest value of profit opportunities 
    function maxProfit(array) {
	
  	var profitArray = array.map(function(currentValue, index) {
  		return array.slice(index, array.length)
  					.sort(function(a, b){return b-a})[0] - currentValue		
  	})
    
    return profitArray.sort(function(a, b){return b-a})[0];

}

console.log(maxProfit([45, 24, 35, 31, 40, 38, 11]));
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    • 0
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    • To display code include the following tags: 
    
function maxStock(arr){
  if(arr[0] > arr[1]){
    var redundant = arr.shift();
    var newArr = arr;
    var maxPrice = Math.max.apply(null, newArr);
    var minPrice = Math.min.apply(null, newArr);
     if(newArr.indexOf(maxPrice) > newArr.indexOf(minPrice)){
        var maxProfit = maxPrice - minPrice;
        } else {
          while(newArr.indexOf(minPrice) > newArr.indexOf(maxPrice)){
           newArr.splice(newArr.indexOf(minPrice), 1);
          }
           var newMin = Math.min.apply(null, newArr);
           maxProfit = maxPrice - newMin;
        }
  }
   return maxProfit;
 }
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    • 0
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    • To display code include the following tags: 
    function maxStock(arr){
  if(arr[0] > arr[1]){
    var redundant = arr.shift();
    var newArr = arr;
     for(var i = 0; i < newArr.length; i++){
        var maxPrice = Math.max.apply(null, newArr);
        var minPrice = Math.min.apply(null, newArr);
        
        if(newArr.indexOf(maxPrice) > newArr.indexOf(minPrice)){
        var maxProfit = maxPrice - minPrice;
        } else {
          while(newArr.indexOf(minPrice) > newArr.indexOf(maxPrice)){
           newArr.splice(newArr.indexOf(minPrice), 1);
          }
           var newMin = Math.min.apply(null, newArr);
           maxProfit = maxPrice - newMin;
        }
    }
  }
   return maxProfit;
 }
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    • 0
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    • The question is a bit misleading. It says "by buying a stock at the given price and then selling the stock later on". The buy price is not given. Would be much clearer to say "buy once and sell once".
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    • I tried this way... 
    function stock(array){
	
  var min= array[0];
  var max= -1;
  var profit= 0;
  
  for(var idx in array){
    
    if(array[idx]<min){
      min= array[idx];
      max= -1;
    }

    if(array[idx]>max){
      max= array[idx];
      if(max-min>profit){
        profit= max-min;
      }
    }

  }

    return profit;
}

console.log(stock([44, 30, 24, 32, 35, 30, 40, 38, 15]));
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    • I'd like to suggest that using array.reduce() is the most streamlined approach for such a problem: 
    function StockPicker(arr) {
	let trueMax = 0;
	arr.reduce(function(prev, item){
		let max = Math.max(item - prev, 0);
		
		max > trueMax ? trueMax = max : null;
		
		if (max === 0) {
			return item;
		} else {
			return prev;
		}
	});
	
	if (trueMax === 0) {
		return -1;
	}
	return trueMax;
}
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    • -4
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    • My version, but I am not quite convinced by this problem. 
    var maxArray = arr;
  maxArray.shift();
  var minArray = arr;
  minArray.pop();

  var min = Math.min.apply(null, minArray);
  var max = Math.max.apply(null, maxArray);

  return max-min;
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