# Codingame: Stock Exchange Losses

## The Goal

`t0`and by selling it at a later date

`t1`. The loss will be expressed as the difference in value between

`t0`and

`t1`. If there is no loss, the loss will be worth 0.

## Game Input

**Input**

Line 1: the number `n` of stock values available.

Line 2: the stock values arranged in order, from the date of their introduction on the stock market `v _{1}` until the last known value

`v`. The values are integers.

_{n}**Output**

`p`, expressed negatively if there is a loss, otherwise 0.

**Constraints**

`n`< 100000

0 <

`v`< 2

^{31}

## VB.net Solution

Based on the first point p we got, we subtract the second point q. If q is smaller than p, we have a loss. What we now must do is maximizing the loss, so we'll go from point to point and compare the currently highest point p with every new point q. Since we can find a new high with every new point we get, we also have to maximize p. I thought visually about this problem. Look at this graph:

Imagine, you put a straight line between the first and the second point. Now you keep the first point fixed as an anker and rotate the line to the third point. That one is higher than the second, and as such the loss is less than the already known. Now we have a new high and let the third point be the anker and go down to the fourth and fifth. We now found a new maximum loss. With the next point we have to move the anker point again and the game goes on. When implemented, it's as simple as the following VB.net snippet:

Module Solution Sub Main () Dim n as Integer n = Console.ReadLine() Dim vs(n) as String vs = Console.ReadLine().split(" ") Dim p as Integer Dim d as Integer p = Convert.toInt32(vs(0)) d = 0 For Each v As String In vs Dim t as Integer t = Convert.toInt32(v) d = Math.max(d, p - t) ' max loss p = Math.max(p, t) ' max seen value Next Console.WriteLine(- d) End Sub End Module

It's also possible to reverse the idea so that the negation at the end isn't necessary:

Module Solution Sub Main () Dim n as Integer n = Console.ReadLine() Dim vs(n) as String vs = Console.ReadLine().split(" ") Dim p as Integer Dim d as Integer p = Convert.toInt32(vs(0)) d = 0 For Each v As String In vs Dim t as Integer t = Convert.toInt32(v) d = Math.min(d, t - p) ' max loss p = Math.max(p, t) ' max seen value Next Console.WriteLine(d) End Sub End Module

Alternatively, a two-pass solution is possible by finding the index of the minimum element first and then find the maximum up to the index of the minimum element. The maximum loss would be the difference between this minimum and maximum. However, the one-pass solution described above is much more efficient.