在股票价值数组中查找买入/卖出价格，以最大化正差

Question

今天在面试中遇到这个问题，它的优化解决方案让我感到冷淡（这很打击，因为我真的很想为这家公司工作......）

给定一个实际值数组，每个数组都代表股票价值在任意一段时间后，找到一家公司的最佳买入价及其相应的最佳卖出价（低买高卖）。

为了举例说明，让我们以 Z 公司的股票代码为例

55.39 109.23 48.29 81.59 105.53 94.45 12.24

：请注意，数组是按时间“排序”的，即随着时间的推移，值会附加到数组的右端。因此，我们的买入价值将（必须）位于卖出价值的左侧。

（在上面的示例中，理想的解决方案是以 48.29 买入并以 105.53 卖出）

我用 O(n² ）复杂性（用java实现）：

// returns a 2-element array: first element is the index in the argument array
// of the best buying price, and the second element is the index of the best
// selling price which, collectively, maximize the trading return
//
// if there is no favorable trading (e.g. prices monotonically fall), null is returned
public int[] maximizeReturn(ArrayList<Double> prices) {
  int [] retval = new int[2];
  int BUY = 0, SELL = 1;
  retval[BUY] = retval[SELL] = -1; // indices of buy and sell prices, respectively

  for (int i = 0; i < prices.size(); i++) {
    for (int j = i + 1; j < prices.size(); j++) {
      double difference = prices.get(j).doubleValue() - 
                          prices.get(i).doubleValue();

      if (difference > 0.0) {
        if (retval[BUY] < 0 || difference > prices.get(retval[SELL]).doubleValue() - 
                                            prices.get(retval[BUY]).doubleValue()) {
          retval[BUY] = i;
          retval[SELL] = j;
        }
      }
    }
  }
  return (retval[BUY] > 0 ? retval : null);
}

这就是我搞砸的地方：有一个线性时间O(n)解决方案，我在试图弄清楚它时完全失败了（是的，我知道， 失败）。有谁知道如何实现线性时间解决方案？ （任何您熟悉的语言）谢谢！

编辑

我想，对于任何感兴趣的人来说，我今天刚刚收到消息，我没有得到我面试的工作，他们问我这个问题。 :(

Answer 1

在 C# 中：

static void Main(string[] args)
{
    double[] values = new double[7]{55.39, 109.23, 48.29, 81.59, 105.53, 94.45, 12.24};

    double max = double.MinValue, maxDiff = double.MinValue, diff = 0;

    for (int i = 1; i < values.Length; i++)
    {
        if (values[i] > values[i - 1])
        {
            //trending upward, append to existing differential
            diff += values[i] - values[i - 1];
        }
        else
        {
            //trending downward, reset the diff
            diff = 0;
        }

        if (diff > maxDiff)
        {
            maxDiff = diff;
            max = values[i];
        }
    }

    Console.WriteLine("Buy at {0}; Sell at {1}", max - maxDiff, max);
}

编辑：基于 @Joe 失败测试用例的新算法 - Nice Catch BTW！这也与@Doug T 现在的答案相同......

static void Main(string[] args)
{
    double[] values = new double[8] { 55.39, 109.23, 48.29, 81.59, 81.58, 105.53, 94.45, 12.24 };

    double max = double.MinValue, maxDiff = double.MinValue, diff = 0;
    double bottom = values[0];

    for (int i = 1; i < values.Length; i++)
    {
        diff += values[i] - values[i - 1];

        if (diff > maxDiff)
        {
            maxDiff = diff;
            max = values[i];
        }

        if (values[i] < bottom)
        {
            bottom = values[i];
            diff = 0;
        }
    }

    Console.WriteLine("Buy at {0}; Sell at {1}", max - maxDiff, max);
}

Answer 2

这是一个尝试（C++）。基本上每次我追踪一个新的顶部时，我都会尝试看看这是否是迄今为止最好的利润。我知道，“底部”肯定早就被发现了。那时我记得顶部、底部和当前的最大利润。如果稍后发现新的底部，那么它是在当前顶部之后，因此我们必须重置顶部，看看稍微较低的“顶部”是否可以产生更好的利润。

#include <iostream>

int main()
{

    double REALLY_BIG_NO = 1e99;
    double bottom = REALLY_BIG_NO; // arbirtrary large number
    double currBestBuy = 0.0;
    double top = 0.0;
    double currBestSell = 0.0;
    double profit = 0.0;

    // array of prices
    double prices[] = {10.50, 55.39, 109.23, 48.29, 81.59, 105.53, 94.45, 12.24, 152.0, 2, 170.0};
    int numPrices = 10;// number of prices

    for (int i = 0; i < numPrices; ++i)
    {
         if (prices[i] < bottom)
         {
            bottom = prices[i];
            // reset the search on a new bottom
            top = 0.0;
         }
         else if (prices[i] > top)
         {
            top = prices[i];
           // calculate profit
            double potentialProfit = (top - bottom);
            if (potentialProfit > profit &&
                bottom != REALLY_BIG_NO)
            {
                profit = potentialProfit;
                currBestSell = top;
                currBestBuy = bottom;
            }
         }
    }

    std::cout << "Best Buy: " << currBestBuy << "Best Sell: " << currBestSell << std::endl;
}

到目前为止，我已经使用了一堆不同的输入集，到目前为止我还没有遇到任何问题...（如果您测试这个并发现任何错误，请告诉我）

我强烈建议使用 Austin Salonen 对此问题的更新答案并根据您的语言进行调整。

Answer 3

这个想法很简单。保留两个指针，lo 和 hi。
执行 Foor 循环

1. 如果价格高于 hi，则更新 hi = 价格，
2. 如果价格低于 hi，则继续。那么 lo 和 hi 就是可能的候选者之一。计算利润，如果比之前的利润大则将其存储并重置 lo, hi 为价格

def getBestProfit(prices):
    lo = hi = profit = 0


    for price in prices:
        if lo == 0 and hi == 0:
            lo = hi = price

        if price > hi:
            hi = price

        if price < low:
            tmpProfit = hi - lo
            if tmpProfit > profit:
                profit = tmpProfit

            lo = hi = price
    return profit

就这样

Answer 4

void getBestTime (int stocks[], int sz, int &buy, int &sell){
int min = 0;
int maxDiff = 0;
buy = sell = 0;
for (int i = 0; i < sz; i++) 
{
    if (stocks[i] < stocks[min])
    {
        min = i;
    }
    int diff = stocks[i] - stocks[min];
    if (diff > maxDiff) 
    {
        buy = min;
        sell = i;
        maxDiff = diff;
    }
}}

以防万一您更喜欢这个答案。我在另一个网上找到了它，但仍然如此。
来源：http://leetcode.com/2010 /11/最佳买入和卖出股票时间.html

Answer 5

      public void profit(float stock[], int arlen ){
            float buy = stock[0];
            float sell = stock[arlen-1];
            int bi = 0;
            int si = arlen - 1;

            for( int i = 0; i < arlen && bi < si ; i++){

                    if( stock[i] <  buy && i < si){
                            buy = stock[i];
                            bi = i;
                    }
                    if(stock[arlen - i - 1] > sell &&  (arlen - i -1)  > bi){
                            sell = stock[arlen - i - 1];
                            si = arlen - i - 1;
                    }
            }
            System.out.println(buy+" "+sell);
    }

Answer 6

我真的必须指出，作为一个面试问题，希望你能解决它，因为 O(n) 是近乎荒谬的。面试问题旨在证明您可以解决问题，并且您能够解决它。事实上你用 O(N^2) 与 O(N) 解决它的事实应该是无关紧要的。如果一家公司因为不能在 O(N) 时间内解决这个问题而放弃雇用你，那么无论如何，这可能都不是你想要工作的公司。

Answer 7

我想描述一下我是如何解决这个问题的，以便更容易理解我的代码：

（1）对于每一天，如果我必须在那天出售我的股票，我可以支付的最低购买金额是多少它？本质上，我会跟踪那一天之前的最低价格

(2) 对于每一天，如果我在那天出售，我会赚多少钱？ （当天的股价 - 最低价格）

这表明我必须跟踪两件事：（1）迄今为止的最低股价（2）迄今为止的最佳收益。

问题变成选择哪一天出售。我会在能给我带来最好收益的那一天出售。这是我的Java代码：

    public static void findBestDeal(double [] stocks) {
    double minsofar = stocks[0];
    double bestsofar = 0.0;

    for(int i=1; i< stocks.length; i++) {

        // What is the cheapest price to buy it if I'm going to sell it today
        if(stocks[i-1] < minsofar) {
            minsofar = stocks[i-1];
        }

        // How much do I earn if I sell it on ith day?
        double current_deal = stocks[i] - minsofar;

        // Is selling today better?
        if(current_deal > bestsofar) {
            bestsofar = current_deal;
        }
    }

    System.out.println("Found the best deal: " + bestsofar + " (Bought at " + minsofar + " and sold at " + (minsofar+bestsofar) + ")");

}

Answer 8

这是我的 O(n) 实现。我正在使用更改数组来计算最大利润以及买卖日期。
欢迎您对此发表评论。

#include<stdafx.h>
#include<stdio.h>

int main()
{
    //int arr[10] = {15, 3, 5,9,10,1,6,4,7,2};
    int arr[7] = {55.39, 109.23, 48.29, 81.59, 105.53, 94.45, 12.24};
    int change[7];
    int n=7;
    for(int i=1;i<=n;i++)
    {
    change[i] = arr[i]- arr[i-1];
    }
    int i=0,index = 0;
    int sum = 0;
    int maxsum = 0;
    int startpos = 0;
    int endpos = 0;
    while(index < n)
    {
        sum = sum + change[index];
        if(maxsum < sum)
        {
        maxsum = sum; 
        startpos = i;
        endpos = index;

        }
        else if (sum<0) // negative number ,set sum to zero
        {
        sum = 0;
        i=index+1;
        }
        index++;
    }

    printf("max profit is%d %d %d", maxsum , startpos, endpos+1 );
}

Answer 9

在我努力学习 Go 的过程中，也为了在这方面绞尽脑汁，这是我的尝试。

func GetMaxProfit2(prices []float64) (float64, float64) {
    var min, max, pmin, pmax int

    for i, v := range prices {
        if v - prices[min] > prices[max] - prices[min] {
            pmax = max
            max = i
        }
        // Reset the max when min is updated.
        if v < prices[min] {
            pmin = min
            min = i
            pmax = max
            max = i
        }
    }

    // If min is ahead of max, reset the values back    
    if min >= max {
        min = pmin
        max = pmax
    }

    return prices[min], prices[max]
}

Answer 10

这是我使用 Javascript 的尝试。该脚本以 O(N) 计算答案：

//Main Stock Array
var stock = [15, 20, 0, 3, 30, 45, 67, 92, 1, 4, 99];


//Setup initial variable state
var ans = {}, tmp = {}; //These are just for namespacing / syntatic sugar
ans.minVal = stock[0];
ans.minInd = 0;
ans.maxDiff = stock[1] - stock[0];
ans.maxInd = 1;
tmp.minInd = ans.minInd;
tmp.minVal = ans.minVal;

//Basically we iterate throught the array. If we find a new low, we start tracking it. Otherwise we compare the current index against the previously found low
for(i = 1; i <= stock.length-1; i++) {
    if(tmp.minVal > stock[i]) {
        tmp.minVal = stock[i];
        tmp.minInd = i;
    } else {
        ans.diff = stock[i] - stock[tmp.minInd];
        if(ans.diff > ans.maxDiff) { //Looks like we found a new maxDifference. Lets log the indexes
            ans.maxDiff = ans.diff;
            ans.maxInd = i;
            ans.minInd = tmp.minInd;
            ans.minVal = tmp.minVal;
        }
    }
}

document.write('You should buy your stocks on day ' + ans.minInd + ' and sell on day ' + ans.maxInd);

Answer 11

这是一个实际有效的 C 解决方案：

void bestBuySell()
{
双倍价格[] = {10.50, 10.0, 3.0, 194.0, 55.39, 2.0, 109.23, 48.29, 81.59, 105.53, 94.45, 191.0, 200.0, 12.24};
int 数组大小 = 14;
double bestBuy = 价格[0]，bestSell = 价格[1]，bestPotentialBuy = 价格[0]；
双潜在利润 = 价格[1] - 价格[0]；

for(int i = 1; i < (arrSize-1); i++)
{
    if(prices[i] < bestBuy)
        bestPotentialBuy = prices[i];            

    if((prices[i+1] - bestPotentialBuy) > potentialProfit)
    {
        bestBuy = bestPotentialBuy;
        bestSell = prices[i+1];
        potentialProfit = prices[i+1] - bestPotentialBuy;
    }
}

printf( "bestBuy %f bestSell %f\n", bestBuy, bestSell );

}

Answer 12

1.我们不能简单地将数值中最小的金额作为“Best Buy”，最大的金额作为“Best Sell”，因为“Sell”必须在“Buy”之后发生。

2.我们不能将记录的最低值视为“百思买”，因为随后几天的股票价值可能与记录的最低值的差异可能产生的利润可能低于“记录的利润”。

3.Best Buy 和 Best Sell 被视为单一变量，因为正是这些值之间的正差才能产生最大利润。

4.由于过去任何记录的最低价都是潜在的买入候选者，因此必须始终根据记录的最低价和当天的股价检查最大利润条件。因此，我们始终必须跟踪记录的最低价，但只是存在由于原因 2，记录的最小值不构成“百思买”。

现在，下面的代码在 O(n) 次内执行将有意义。

public class BestStockBuyAndSell {

public static void main(String[] args) {

    double[] stockPrices = {55.39,109.23,48.29,81.59,105.53,94.45,12.24};
    int [] bestBuySellIndex = maxProfit(stockPrices);

    System.out.println("Best Buy At "+stockPrices[bestBuySellIndex[0]]);
    System.out.println("Best Sell At "+stockPrices[bestBuySellIndex[1]]);

    System.out.println("Max Profit = "+(stockPrices[bestBuySellIndex[1]]-stockPrices[bestBuySellIndex[0]]));

}

public static int[] maxProfit(double[] stockPrices)
{
    int bestBuy=0;
    int bestSell=0;

    int[] bestCombination ={bestBuy,bestSell};
    double recordedMinimum = stockPrices[bestBuy];
    int recordedMinimuIndex = bestBuy;
    double bestProfitSofar = stockPrices[bestSell] - stockPrices[bestBuy];

    for(int i=1;i<stockPrices.length;i++)
    {
        if(stockPrices[i] - recordedMinimum > bestProfitSofar)
        {

            bestProfitSofar = stockPrices[i] - recordedMinimum;
            bestSell = i;
            bestBuy = recordedMinimuIndex;
        }

        if(stockPrices[i] < recordedMinimum)
        {
            recordedMinimuIndex = i;
            recordedMinimum = stockPrices[i];
        }

    }

    bestCombination[0] = bestBuy;
    bestCombination[1] = bestSell;


    return bestCombination;

}

}

Answer 13

我针对这个问题提出了以下算法，似乎适用于所有输入。另外，如果股票价值持续下跌，程序会输出不购买该股票：

  public class GetMaxProfit 
  { 

  double minValue = -1, maxValue = -1;
  double maxDiff = 0;

  public void getProfit(double [] inputArray){
    int i=0, j=1;
    double newDiff = 0;
    while(j<inputArray.length){
         newDiff = inputArray[j]-inputArray[i];
         if(newDiff > 0){
             if(newDiff > this.maxDiff){
               this.minValue = inputArray[i];
               this.maxValue = inputArray[j];
               this.maxDiff = newDiff;
             }
        }
        else{
            i = j;
        }
        j++;
    }
 }

 public static void main(String[] args) {
    // TODO Auto-generated method stub
    GetMaxProfit obj = new GetMaxProfit();

    obj.getProfit(new double[]{55.39, 19.23, 14.29, 11.59, 10.53, 9.45, 1.24});
    if(obj.minValue != -1 && obj.maxValue != -1){
      System.out.println("Buy Value for the input: "+obj.minValue);
      System.out.println("Sell Value for the input: "+obj.maxValue);
      System.out.println("Best profit for the input: "+obj.maxDiff);
            }
            else
               System.out.println("Do Not Buy This STOCK!!);

 }

}

这有什么陷阱吗？时间复杂度为O(N)

Answer 14

这是我的解决方案，与@Doug T 相同，但我也在索引中跟踪这一天。请提供反馈。

 int prices[] = {4,4,5,6,2,5,1,1};
 //int prices[] = {100, 180, 260, 310, 40, 535, 695};

 int currentBestSellPrice=0;
 int currentBestBuyPrice=0;
 int lowindex=0;
 int highindex=0;
 int low=prices[0];
 int high=prices[0];
 int profit=0;
 int templowindex=0;
 for(int i=0; i< prices.length;i++)
 {
     // buy low
     if(prices[i] < low && i+1 < prices.length)
     {
         low = prices[i];  
         templowindex=i;
         high=0;
     }
     // sell high
     else if(prices[i] > high)
     {
         high = prices[i];
         int potentialprofit = (high-low);
         if(potentialprofit > profit)
         {
             profit = potentialprofit;
             currentBestSellPrice = high;
             currentBestBuyPrice = low;
             highindex=i;
             lowindex=templowindex;
         }
     }
 }


 System.out.println("Best Buy Price : "+ currentBestBuyPrice + " on day "+ lowindex);
 System.out.println("Best Sell Price : "+ currentBestSellPrice+ " on day "+ highindex );

Answer 15

F# 解决方案适合那些对功能感兴趣的人。我不会说，虽然有那么大的不同。

let start, _, profit = 
    [55.39; 109.23; 48.29; 81.59; 81.58; 105.53; 94.45; 12.24 ]
    |> Seq.fold (fun (start,newStart,profit) i -> 
                    let start = defaultArg start i
                    let newStart = defaultArg newStart i
                    let newProfit = i - newStart
                    if profit < newProfit 
                    then  Some newStart, Some newStart,newProfit
                    else if start > i 
                    then Some start, Some i, profit 
                    else Some start,Some newStart,profit) (None,None, 0.0)
printf "Best buy: %f; Best sell: %f" start.Value (start.Value + profit)

输出：

Best buy: 48.290000; Best sell: 105.530000

Answer 16

这是我在 Ruby 中的解决方案：

values = [55.39, 109.23, 48.29, 81.59, 105.53, 94.45, 12.24]

max_diff = 0
diff = 0
min = values[0]
max = 0

values.each_with_index do |value, index = 1|
  # get an array of the previous values before the current one
  lag_values = values[0..index]

  # get the minimum of those previous values
  min_lag_value = lag_values.min

  # difference between current value and minimum of previous ones
  diff = values[index].to_i - min_lag_value.to_i

  # if current difference is > previous max difference, then set new values for min, max_diff, and max
  if diff > max_diff
    max_diff = diff
    min = min_lag_value
    max = values[index]
  end
end

min # => 48.29
max # => 105.3
max_diff # => 57

干杯

Answer 17

我得到了 100% 的同样的结果，给你。

public int solution(int[] A) {
      if (A == null || A.length<=1){
            return 0;
        }
        int minValue = Math.min(A[0], A[1]);
        int profit = A[1] - A[0];
        for (int i = 2; i < A.length; i++) {
          minValue = Math.min(minValue, A[i]);
          profit = Math.max(A[i] - minValue,profit);
        }

        return profit > 0 ? profit : 0;
}

Answer 18

我的想法是，对于每个指数i，出售这只股票的理想指数是什么？这当然是i之后最大值的索引。这将我们的问题简化为在 [1 ... n] 中为每个 i 找到索引 i 之后的最大元素如果我们可以这样做O(n) 时间，然后我们可以在其中找到最佳选择并报告它。

一种方法是从数组末尾开始遍历，维护两个变量，一个保存我们迄今为止遇到的最大元素max_till_now，一个保存我们可以获得的最大利润到目前为止，盈利。这只是为了让我们可以一次性完成此操作。我们还可以使用额外的空间，对于每个元素 i，存储其范围 [i + 1 ... n] 中最大元素的索引，然后查找最大利润。

这是我的Python代码：

def buyLowSellHigh(L):
    length = len(L)
    profit = 0
    max_till_now = L[length - 1]
    for i in xrange(length - 2, -1, -1):
        if L[i] > max_till_now: max_till_now = L[i]
        else:
            if max_till_now - L[i] > profit: profit = max_till_now - L[i]
    return profit

Answer 19

另一个 Ruby 解决方案：

# Here's some examples. Please feel free to give your new test.
values = [55.39, 109.23, 48.29, 81.59, 105.53, 94.45, 12.24]
# values = [5, 6, 4, 7, 9, 8, 8]
# values = [5, 10, 4, 6, 7]
# values = [5, 10, 4, 6, 12]
# values = [1, 2, 3, 4, 5]



# Initialize parameters.
min = values[0]
best_buy_time = values[0]
best_sell_time = values[0]
max_profit = 0



# This solution is based on comparing previous k elements and k+1 one.
# The runtime is O(n) and it only use O(1) auxiliary storage.
values.each_with_index do |value, index = 1|

  # Check value in this turn.
  puts value

  # Check current value is bigger than min or not.
  # If not, we find the new min.
  if value <= min
    min = value

  # If current value is bigger than min and
  # (value - min) is bigger than previous max_profit,
  # set new best_buy_time, best_sell_time & max_profit.
  else
    if value - min >= max_profit
      best_buy_time = min
      best_sell_time = value
      max_profit = value - min
    end

  end

end



# Let's see about the result.
puts "\nbest_buy_time: ", best_buy_time, "\nbest_sell_time: ", best_sell_time, "\nmax_profit: ", max_profit

Answer 20

那这个呢？

min = 100000000
max = 0

for elem in inp:
    if elem < min:
       min = elem
    tempMax = elem-min
    if tempMax > max:
        max = tempMax

print(max)

Answer 21

JavaScript 中的解决方案

var stockArr = [13931, 9889, 987, 4, 89, 100];

function getBestTime(sortedArr) {
  var min = 0;
  var buyIndx = 0;
  var saleIndx = 0;
  var maxDiff = 0;
  for (var i = 0; i < stockArr.length; i++) {
    if (stockArr[i] < stockArr[min]) {
      min = i;
    }
    var diff = stockArr[i] - stockArr[min];
    if (diff > maxDiff) {
      buy = min;
      sale = i;
      maxDiff = diff;
    }
  }
  return {
    buy:buy+1,
    sale:sale+1,
    diff:maxDiff
  }
}

console.log(getBestTime(stockArr));

Answer 22

这是一个 javascript 解决方案。

function getMax(arr){
        //we need more than at least 3 ints to be able to do this
        if(arr.length <= 1) return arr;
        // get the minimum price we can sell at to make a profit
        var min = arr[0];
        //get the first potential maximum profit
        var max = arr[1] - arr[0];

        //while looping through we must get a potential value, 
       //we can then compare that using the math.max using the maximum
      //and the potential prices that we have seen. Once the loop runs the ouput here should be 6!
        for(var i = 1; i < arr.length; ++i){
            var current = arr[i];
            var potential = current - min;

            max = Math.max(max, potential);
            min = Math.min(min, current);
        }

        return max;
    }

    console.log(getMax([10, 7, 5, 8, 11, 9]));

运行时间是 O(n)

Answer 23

scala 中的解决方案：

示例： [ 7, 2, 5, 6, 1, 3, 6, 4 ]

保留指向最后一个最低股票价格（lastStockPrice）的指针，并将其与当前股票价格进行比较。当你到达当前股价<最后最低股票价格，您更新lastStockPrice。

在循环数组时，跟踪 currentPrice 和 lastStockPrice 之间的最大差异（利润），因为当您更新 lastStockPrice 时，利润可能会发生变化。

下面的 scala 代码在 O(n) 时间内运行并占用恒定的空间。

object Solution {
    def maxProfit(prices: Array[Int]): Int = {
        var lastStockPrice = Int.MaxValue
        var maxProfit = 0
        for(currentPrice <- prices){
            if(currentPrice < lastStockPrice){
                lastStockPrice = currentPrice;
            }else if(currentPrice - lastStockPrice > maxProfit){
                maxProfit = currentPrice - lastStockPrice;
            }
        }
        maxProfit
    }
}

Answer 24

解决这个问题的逻辑与使用 Kadane 算法的“最大子数组问题”相同。由于到目前为止还没有人提到这一点，我认为每个人都知道这是一件好事。

所有直接的解决方案都应该有效，但是如果面试官通过给出价格差异数组来稍微扭曲问题，例如：对于 {1, 7, 4, 11}，如果他给出 {0, 6, -3, 7} ，您可能最终会感到困惑。

这里的逻辑是计算原始数组的差值(maxCur +=prices[i]-prices[i-1])，并找到给出最大利润的连续子数组。如果差值低于 0，则将其重置为零。

class Solution:
def maxProfit(self, prices: List[int]) -> int:
    
    _currmax = 0
    _globalMax = 0
    
    for i in range(1,len(prices)):
        
        _currmax = max(_currmax+(prices[i]-prices[i-1]),0)
        _globalMax = max(_globalMax,_currmax)
        
    return _globalMax