## Hackerrank - Stock Maximize Solution

Your algorithms have become so good at predicting the market that you now know what the share price of Wooden Orange Toothpicks Inc. (WOT) will be for the next number of days.

Each day, you can either buy one share of WOT, sell any number of shares of WOT that you own, or not make any transaction at all. What is the maximum profit you can obtain with an optimum trading strategy?

For example, if you know that prices for the next two days are , you should buy one share day one, and sell it day two for a profit of . If they are instead , no profit can be made so you don't buy or sell stock those days.

**Function Description**

Complete the *stockmax* function in the editor below. It must return an integer that represents the maximum profit achievable.

stockmax has the following parameter(s):

*prices*: an array of integers that represent predicted daily stock prices

**Input Format**

The first line contains the number of test cases .

Each of the next pairs of lines contain:

- The first line contains an integer , the number of predicted prices for WOT.

- The next line contains n space-separated integers , each a predicted stock price for day .

**Constraints**

**Output Format**

Output lines, each containing the maximum profit which can be obtained for the corresponding test case.

**Sample Input**

```
3
3
5 3 2
3
1 2 100
4
1 3 1 2
```

**Sample Output**

```
0
197
3
```

**Explanation**

For the first case, you cannot obtain any profit because the share price never rises.

For the second case, you can buy one share on the first two days and sell both of them on the third day.

For the third case, you can buy one share on day 1, sell one on day 2, buy one share on day 3, and sell one share on day 4.

### Solution in Python

```
def stockmax(prices):
m = prices.pop()
maxsum = 0
arrsum = 0
for p in reversed(prices):
m = max(m, p)
maxsum+=m
arrsum+=p
return maxsum-arrsum
for _ in range(int(input())):
n = int(input())
prices = list(map(int,input().split()))
print(stockmax(prices))
```

*Similar alternative solution*

```
from collections import Counter
def stockmax(prices):
m = prices.pop()
maxarr = Counter()
for i in range(len(prices)-1, -1,-1):
m = max(m, prices[i])
maxarr[i] = m
return sum(maxarr[i]-v for i,v in enumerate(prices))
for _ in range(int(input())):
n = int(input())
prices = list(map(int,input().split()))
print(stockmax(prices))
```

*Logic*

Let us assume we have the following prices

`1 3 1 2`

Our goal is to create another array which holds maximum price of the future (including current day, we can buy and sell the same day. If future prices are smaller)

```
Day 1 -> Maximum future price = 3
Day 2 -> Maximum future price = 3
Day 3 -> Maximum future price = 2
Day 4 -> Maximum future price = 2
```

We can do this by iterating from backwards.

```
Day 4 -> m = 2
Day 3 -> 1 < 2 -> m = 2
Day 2 -> 3 > 2 -> m = 3
Day 1 -> 1 < 3 -> m = 3
```

So our maxarr will hold these value

```
>>> maxarr
[3,3,2,2]
```

Now we just have to sum the difference between the values in maxarr and prices list

```
>>> prices
[1,3,1,2
>>> maxarr
[3,3,2,2]
```

```
[3-1, 3-3, 2-1, 2-2]
[2, 0, 1, 0]
Total profit = 2 + 1 =3
```