# Vector thinking for Max Profit

Occasionally I am lucky enough to run into a problem that doesn’t already have many vector solutions, so I can demonstrate how vector thinking can allow you to find an elegant solution. To build up to my insight let’s solve 2 easy variations and then we can get to the two more interesting puzzles.
The problem in question is called Best Time to Buy and Sell or more popularly simply Max Profit.
The theme is always the same you are given an ordered list of positive numbers that represent the evolution of the price of the security. You must find the maximum profit you could have made given this price series. You can only hold 1 unit at a time and you can not short, which means you must buy before you can sell (you can only profit from the price evolving upward).
The simplest and most well know variation is the 1 transaction problem (leetcode problem 121): you can only buy and sell at most 1 time what is the max profit that can be earned from the following series:
example series:
3 2 5 1 3 2 4 9 5 12 3 5
The key insight into this problem is that we are looking for the largest difference between the current element and the smallest element we have seen.
To motivate this insight, let’s look at ever larger versions of this problem starting with just one price to detect the pattern.
3 -> 0
Given just one price we can only buy and sell on the same day, so effectively max profit is 0
3 2 -> 0
Given these two points we can still only earn 0, since we must sell after we buy.
3 2 5 -> 3
Now that we have a higher number after a lower number we can plainly see that of the two options buy at 3 or buy at 2, buy at 2 and sell at 5 is better.
3 2 5 1 -> 3
The answer is still 3, since we can’t sell higher than 5, and even if we bought at 1 at the end there are no days left to sell
3 2 5 1 3 -> 3
We still do best by buying at 2 and selling at 5, buying at 1 and selling at 3 only earns 2.
3 2 5 1 3 2 -> 3
Plainly the 2 at the end doesn’t improve
3 2 5 1 3 2 4 -> 3
We can now either buy at 2 and sell at 5 or buy at 1 and sell at 4, but our max profit is still 3.
3 2 5 1 3 2 4 9 -> 8
Finally, the max profit changes! We can now buy at 1 and sell at 9. As new elements are added the new reward will be a function of the lowest element seen thus far.
We then get our solution:
{max x-mins x} 3 2 5 1 3 2 4 9 5 12 3 5
11, where we buy at 1 and sell at 12.
Or in python we could say something like:

```import itertools
from operator import sub
def max_profit(p):
return max(map(sub,p,itertools.accumulate(p,min)))
```

To unpack this a bit, mins returns the min element up to that point:
mins[3 2 5 1 3 2 4 9 5 12 3 5]
3 2 2 1 1 1 1 1 1 1 1 1
we can then do element-wise subtraction to see what the profit is from selling at each point
0 0 3 0 2 1 3 8 4 11 2 4
taking the running max we get the max profit so far,
0 0 3 3 3 3 3 8 8 11 11 11
we can see that this series is exactly the results that we got when we were looking at the max profit for the prefixes of the series.

Easy enough, let’s look at version 2 of this problem (leetcode 122):
unlimited number of transactions: We are now allowed to buy and sell as often as we like provided we never own more than 1 share of the stock and we still must sell after we bought. To make things easier we can assume that we can buy and sell on the same day, in practice this doesn’t matter as any day we do this we can consider that we did nothing.
Keeping the same price series as before:
3 2 5 1 3 2 4 9 5 12 3 5
We now notice that we should take advantage of every single time the price goes up. Again let’s look at a smaller example to get some intuition.
3 2 5 -> 3
Nothing fancy here, we buy at 2 and sell at 5, purchasing at 3 doesn’t improve our profit since selling at 2 would incur a loss.
3 2 5 1 3 -> 5
buy at 2 sell at 5, buy at 1 sell at 3
3 2 5 1 3 2 4 -> 7
just add one more purchase at 2 and sell at 4,
3 2 5 1 3 2 4 9 -> 12
Here it becomes interesting, we can look at two interpretations
buy at 2 sell at 5 (3),buy at 1 sell at 3 (2), buy 2 sell at 9 (7) for a total of 12
or we can say:
buy at 2 sell at 5 (3),buy at 1 sell at 3 (2), buy at 2 sell at 4 (2),buy at 4 sell at 9 (5) for a total of 12.
The two approaches are identical since addition commutes, it doesn’t matter how you get from 2 – 9 you will always earn 7.
Which means that we can simply add up the positive steps in the price series. That will be the maximum profit for an unlimited number of transactions:
Our code is simply:
{(x>0) wsum x:1 _ deltas x} 3 2 5 1 3 2 4 9 5 12 3 5
21
Deltas gives us the adjacent differences. We drop the first delta since we cannot sell before we have bought. If the delta is positive >0 we want to include it in our sum. Using (w)eighted(sum) we weight the >0 with a 1 and less then 0 or =0 are given a weight of 0.

Now that we covered the case with 1 transaction and unlimited transactions, we should feel confident that we can tackle 2 transactions. Lucky for us, that’s exactly version III of the problem (leetcode 123):
Getting insight into how to solve this variation should start with thinking about how we could solve this naively. The correct heuristic to reach for is divide and conquer. Suppose that for every step along the price evolution we knew what the max profit was before and the max profit was after. Then we could sum those two max profits and take the largest combination.

assume max_profit function as before:

```r=range(len(p))
max([max_profit(p[0:i])+max_profit(p[i:]) for i in r)])
```

For our example this gives us 15.

What we might notice is that we are recomputing the max profit over for larger and larger prefixes, and for smaller and smaller suffixes.
We know that the prefixes were already computed simply by taking the rolling max instead of the overall max in max_profit. The only thing we’re missing is how to calculate the max_profit of suffixes.
Now we notice a symmetry – while the prefixes are governed by the rolling minimum, the suffixes are bounded by the rolling maximum from the left, i.e. the largest element available at the end to sell into.

To get a sense of this, look at the solutions to the suffixes
3 5 -> 2 (buy at 3 sell at 5)
12 3 5 -> 2 (buy at 3 sell at 5)
5 12 3 5 -> 7 (buy at 5 sell at 12)
9 5 12 3 5 -> 7 (buy at 5 sell at 12)
4 9 5 12 3 5 -> 8 (buy at 4 sell at 12)
Which leads to this solution in q:

``{max reverse[maxs maxs[x]-x:reverse x]+maxs x-mins x}``

maxs x -mins x / is familiar from earlier
If we want to walk backwards through a list, we can reverse the list and apply the logic forwards then reverse the result, which is the same as applying the logic to the suffixes.
So all we need to do is to subtract each element from the running maximum then take the running max of this result to get the max profit for the suffix so far.
In Python this simply becomes:

```from itertools import accumulate as acc
def max_profit_2trans(p):
before=list(acc(map(sub,p,acc( p,min)),max))
p.reverse()
after=list(acc(map(sub,acc(p,max),p),max))
after.reverse()
```

Alright, finally we should be able to tackle the most general version of this problem (leetcode 188). You are given both a series of prices and allowed to make up to k transactions. Obviously, if k is 1, we solved that first. We just solved k is 2. You can verify for yourself, but if k is larger than half the length of the price series the max is the same as the second problem we solved. Since every pair of days can allow for at most one transaction, effectively k is unlimited at that point.
We need to solve the k>2 but less than k<n/2. The standard CS technique that comes to mind is some kind of dynamic programming approach. To quote MIT’s Erik Demaine CS6006 dynamic programing is recursion plus memoization.
Let’s setup the recursive solution and assume we are at a particular (i)ndex in our price series, with k transactions left.
If k equals 0 we return 0
If i is greater than the last index, i.e. there are no elements left in the list: we return 0.
Otherwise the solution to the problem is simply the maximum of 2 options:

do nothing at this step:

0+the function increment i

do something:

If we are (h)olding a share we can sell at this step which adds the current price +the result of this function with one less k and i incremented by 1.
Otherwise we buy at this step, which is minus the current price (we spend money to buy) plus the result of this function with i incremented and we are now holding a share.

Here is the code for this in q:

`p:3 2 5 1 3 2 4 9 5 12 3 5cache:(0b,'0,'til[count p])!count[p]#0f:{[h;k;i]\$[i=count p;0;(h;k;i) in key cache;cache[(h;k;i)];:cache[(h;k;i)]:.z.s[h;k;i+1]|\$[h;p[i]+.z.s[0b;k-1;i+1];.z.s[1b;k;i+1]-p i]]}We can test this and see that it results in the right answer for k=0,1,2f[0b;0;0] -> 0 no surprise with 0 transaction no profit is possiblef[0b;1;0] -> 11 the original problemf[0b;2;0] -> 15 buy at 1, sell at 9, buy at 5 sell at 12`

However, we know that a vector solution is possible for k=1,2,infinity, so we might hope there is a vector solution for when k is 3 and above.
We can analyze the intermediate results of the cache to get some sense of this.

code to look at the table

``````t:(flip `h`k`j!flip key cache)!([]v:value cache)
exec (`\$string[j])!v by k from t where not h ``````

output table:

`````` | 11 10 9 8 7 6  5  4  3  2  1  0
-| --------------------------------
0| 0  0  0 0 0 0  0  0  0  0  0  0
1| 0  2  2 7 7 8  10 10 11 11 11 11
2| 0  2  2 9 9 12 14 14 15 15 15 15
3| 0  2  2 9 9 14 16 16 17 17 18 18
4| 0  2  2 9 9 14 16 16 18 18 20 20``````

What we might notice is that the first row is the running maximum from selling at the prefixes. Then it’s the combination of the previous row and 1 additional transaction. In other words, each row allows you to spend money from the previous row and sell at the current price.
Putting this into action we get the following:

{[k;p]last {maxs x+maxs y-x}[p]/[k;p*0]}

We can look at intermediate rows by writing the scan version of the code and we know it will converge to the unlimited transaction case:

``````{[p]{maxs x+maxs y-x}[p]\[p*0]} p
0 0 0 0 0 0 0 0  0  0  0  0
0 0 3 3 3 3 3 8  8  11 11 11
0 0 3 3 5 5 6 11 11 15 15 15
0 0 3 3 5 5 7 12 12 18 18 18
0 0 3 3 5 5 7 12 12 19 19 20
0 0 3 3 5 5 7 12 12 19 19 21``````

What’s really nice about this code is that it actually describes the problem quite well to the computer and in effect finds exactly what we want.
p*0 / assume we have 0 transactions and therefore 0 dollars at each step.
y-x / subtract the current price(x) from the current profit(y) element wise
maxs y-x/ find the rolling max revenue
x+maxs y-x / add back the current selling price
maxs x + maxs y-x / find the rolling max profit
repeating this k times finds the max profit of the k’th transaction.

Here is the python code that implements this logic:

```import itertools
def max_profit(k: int, prices: List[int]) -> int:
if k==0 or len(prices)<2:
return 0
z=[0]*len(prices)
maxs=lambda x: itertools.accumulate(x,max)
subeach=lambda x,y: map(sub,x,y)
for i in range(k):
return list(z)[-1]
```

I think this post covers this puzzle, let me know if there are interesting variations I haven’t explored.

# Fifo Allocation in KDB

We leave the discussion for more elaborate allocation methods and their respective merits for another time. In this post we will focus on the mechanics of efficiently calculating fifo allocations. Luckily, Jeff Borror covers an elegant implementation in Q for Mortals, which I will only briefly repeat here. The idea is that you can represent the buys and sells as a matrix, where each cell corresponds to the amount allocated to that purchase and sale. By convention rows will correspond to purchases and columns to sales. So in our example, we can write the allocation as

``` | 4 3 2
-| -----
7| 4 3 0
2| 0 0 2
```

I left the corresponding order quantities in the row and columns as headers but they are actually implied. Jeff also gives us the algorithm that produces this matrix.

• First we calculate the rolling sums of the purchase and sales
• 7 9 for purchases
• 4 7 9 for sales
• We then take the cross product minimum
• 4 7 7
4 7 9
• We then take the differences along the columns
• 4 7 7
0 0 2
• We then take the differences along the rows
• 4 3 0
0 0 2
• We are done, as a bonus here is the code in Q
• deltas each deltas sums[buys] &\: sums[sells]

Not only is this rather clever, there is a certain logic that explains how to come to this result. The cumulative sums tells you how much max inventory you have bought or sold till this point. The minimum tells you how much you can allocate so far assuming you haven’t allocated anything. The differences down the columns subtracts the amount you have already allocated to the previous sales. The differences along the rows tells you how much you have already allocated to the previous purchases. Since you can only allocate what hasn’t yet been claimed.

Having read this far, you should feel confident you can easily do fifo allocations in KDB. I know, I did. There are even stack overflow answers that use this method. There is one problem, that occurs the moment you start dealing with a non trivial number of orders. This method uses up n^2 space. We are building a cross product of all the buys and sells. We know that the final matrix will contain mostly zeros, so we should be able to do better. We can use the traditional method for doing fifo allocation. Keep two lists of buys and sells allocated thus far and keep amending the first non zero element of each list and created a list of allocations triplets, (buy, sell, allocated). Although, this is linear implementing this in KDB is rather unKdb like. For incidental reasons, amending data structures repeatedly which is what this algorithm entails is best done by pointers, which in KDB means using globals and pass by reference semantics. It’s not long, but it’s not pretty.

This algorithm, has sufficient performance that we could stop there. However, we could ask is there a way to get all the benefits of the elegant array solution without paying the space cost. The answer is that we can, the trick is that as we have noticed most of the matrix will actually be filled with zeros. In particular, we can see that the matrix will essentially traverse from the top left hand corner to the bottom left hand corner. If we could split the problem into small pieces and then stitch the solutions together we would have the original path of allocations.

I will now briefly sketch out how we can split this problem into pieces and then I will present an annotated version of the code.

If we just look at the first 100 buys and first 100 sells. We can simply apply Jeff’s algorithm. If we wanted to apply it to the next 100 buys and next 100 sells, we would find that we have a problem. We need to know three things, we need to know the index of the buys and sells we are currently up to and any remaining buys and sells that we have not allocated to yet in the previous iteration. Strictly speaking we can only have unallocated quantities on one side, but it is easier to simply keep track of both and letting one list be empty each time.

Here is the annotated code:

This code actually performs much faster than the iterative version and is shorter. This code essentially has no branching except where there are no more allocations on one side and exits early.

If anyone has a better way of computing fifo allocation let me know in the comments.

Below are some timings and graphs of memory usage. The code can be found here.