bruteforce | swiss army nodejs brute force backtester for Gekko trading | Runtime Evironment library

This answer is not React specific, but one approach would be to use array.reduce() to transform each level/node of the structure as shown in the code snippet below.

This seems to be an error message emitted by your program:

Please execute this program with the input file name included as an argument.

We can't know what goes wrong because you don't show the part of your program that emits this.

You say

I have written a program which takes in an single command line argument

but you give multiple arguments:

The problem here is that the way your recursive method works at present isn't amenable to memoization.

It looks like you are using your dp array to store the currently-known minimum count of numbers required to make a total. So dp[5] will be the minimum count of numbers required to make a total of 5. If dp[5] = 3 then you have found a way to make a total of 5 from three numbers, but not yet found a way to make 5 from fewer than three numbers.

Your method shortestSubsequenceWithSum presently returns the minimum count of numbers required to reach the total plus the number of recursive calls currently made. If you want to use memoization you will have to adjust this method to return the minimum count of numbers required to reach the total, regardless of how many levels of recursion there have so far been.

The changes you will need to make are:

• In the handling for the case s == 0, return 0 rather than cnt. This represents being able to make a total of 0 from zero numbers.
• Change the line dp[s] = res to dp[s] = res + 1. res contains the (minimum) count of numbers that make up s - num[i] for each value of i so far, so we add 1 for choosing num[i] in a combination that adds up to s.

These should be enough to get your code working. However, you can in fact move the line dp[s] = res + 1 immediately outside the for loop: we may as well wait for the final value of res before assigning it to dp[s]. You can also remove the parameter cnt from your method shortestSubsequenceWithSum, and all the calls to this method, as this parameter isn't being used any more.

This should leave you with the following:

After seeing your update, this problem is clearly O(a^n)--exponential time--on the surface. This is among the hardest time complexity classes.

Quite frankly, a program written to tackle this problem in a naive manner will, as you suspect, take an incredibly long time. Not just beyond the life span of your computer, as an n of 20 will have a worst-case result on the order 10^100 operations. Even when adjusted to multiple operations per nanosecond, that's an amount of time so unfathomably large that, assuming I'm not calculating wrong here, would well exceed the heat death of the universe. In other words, you will never solve this problem for a sequence of size 20 on an array of size 100,000 using a naive approach.

And with that said, I don't know if a "quick" solution even exists. This appears, at a glance, to be something called the "k-xor problem". Or rather, if we XOR every element in the array with the value of k, then it should be an identical problem. And the "k-xor problem" appears to have such a large importance in cryptography that there have been continued efforts to create quantum algorithms purely to deal with the time complexity issues of the problem. And these solutions are for smaller values of n, like 3 or 4, not 20. Even non-quantum solutions appear to have solutions on the order of time complexity O(a^(n/2)), which means your worst-case runtime complexity is still going to be executing on the order of 10^50 operations, which is still an unfathomably large number.

You say that you're in high school, and considering there is cutting edge research still being done on improving solutions for this kind of problem, I don't think it's reasonable for you to look into efficiency. I would expect an efficient solution to a problem like this to be asked of graduate-level researchers doing thesis work at a minimum, not of a high school student.

Unless there are important details you're leaving out or misunderstanding that could potentially reduce the problem complexity significantly, there's just no way you're going to write an efficient solution. Period.

In short, I would take that as permission to simply ignore the size of the problem and just write a solution that can work on smaller values of a and n. If your teacher somehow knows of a fast algorithm, then maybe they should submit a research paper.

Ok, I ended up solving the redirection issue by using states and hooks let me explain:

I created a new state called redirectState and its setter and a navigate one:

For each position in the combination, the possible values are a subset of the list starting at the position and up to the last k-p-1 element. e.g. for combinations of 6 in 1..100, position 3 can only contain values 3..96

For each of the positon/value pairs, the number of occurrences will be the product of combinations of left side elements and combinations of right side elements.

For example, for combinations of 6 elements within a list of 1..100, the number of times 45 will appear at the third position is the combinations of 2 in 1..44 times the combinations of 3 in 46..100. So we will have C(44,2) * C(55,3) * 45 for that positon/value pair.

You can repeat this calculation for each positon/value pair to obtain a total for each position in the output combinations. Then divide these totals by the number of combinations to get the expected value:

Please let me know if I'm misunderstanding your question.

Binary numbers map one-to-one with decimal numbers, so you can cover a two-character alphabet with binary numbers of length 1, four-characters with length 2, etc.

So for a 40-character alphabet you'll need binary codes of length 6. Then if you have them in a list:

Elements you are trying to access are inside a frame. You have to switch to that frame in order to access these elements.
Try this:

Google eventually treats JavaScript changes to the location when the page loads the same as a server side redirect. Using location changes to canonicalize your URLs will be fine for SEO.

The same caveats that apply to all JavaScript powered pages apply to this case as well:

• Google seems to take longer to index pages and react to changes when it has to render them. JavaScript redirects may no be identified as redirects as quickly as server side redirects would be. It could take Google a few extra weeks or even a couple months longer.
• Google won't see any changes that happen in response to user actions such as clicking or scrolling. Only changes that happen within a couple seconds of page load with no user interaction will be recognized.