algorithm | Algorithms implemented by Java | Learning library

This problem has a fun O(n) solution.

If you draw a graph of cumulative sum vs index, then:

The average value in the subarray between any two indexes is the slope of the line between those points on the graph.

The first highest-average-prefix will end at the point that makes the highest angle from 0. The next highest-average-prefix must then have a smaller average, and it will end at the point that makes the highest angle from the first ending. Continuing to the end of the array, we find that...

These segments of highest average are exactly the segments in the upper convex hull of the cumulative sum graph.

Find these segments using the monotone chain algorithm. Since the points are already sorted, it takes O(n) time.

Below Steps can help to solve issue in chrome 98, for other browser like edge you need to do similar like chrome.

• Requestly with chrome version 98. You need to follow following steps :- Run this command on terminal

  defaults write com.google.Chrome InsecurePrivateNetworkRequestsAllowed -bool true

• Restart your Browser, Not work then restart your machine

• Run 'regedit' to open windows registry (If permission issue came then run that command with Admin command prompt)
• Go to Computer\HKEY_LOCAL_MACHINE\SOFTWARE\Policies\Google\Chrome
• Create new DWORD value with "InsecurePrivateNetworkRequestsAllowed" Name
• Change Value to "1"
• Restart your Browser

Your split splits the list in two ordered halves, so merge consumes its first argument first and then just produces the second half in full. In other words it is equivalent to ++, doing redundant comparisons on the first half which always turn out to be True.

In the true mergesort the merge actually does twice the work on random data because the two parts are not ordered.

The split though spends some work on the partitioning whereas an online bottom-up mergesort would spend no work there at all. But the built-in sort tries to detect ordered runs in the input, and apparently that extra work is not negligible.

Imo, it's a bug in Paramiko. It does not handle correctly absence of server-sig-algs extension on the server side.

Try disabling rsa-sha2-* on Paramiko side altogether:

The reason is this breaking change:

DeflateStream, GZipStream, and CryptoStream diverged from typical Stream.Read and Stream.ReadAsync behavior in two ways:

They didn't complete the read operation until either the buffer passed to the read operation was completely filled or the end of the stream was reached.

And the new behaviour is:

Starting in .NET 6, when Stream.Read or Stream.ReadAsync is called on one of the affected stream types with a buffer of length N, the operation completes when:

At least one byte has been read from the stream, or The underlying stream they wrap returns 0 from a call to its read, indicating no more data is available.

In your case you are affected because of this code in Decrypt method:

At first blush, one would hope this would be enough:

Take a look at the recursive formula for combinations:

Suppose you have a combination space C(n,k). You can divide that space into two subspaces:

• C(n-1,k-1) all combinations, where the first element of the original set (of length n) is present
• C(n-1, k) where first element is not preset

If you have an index X that corresponds to a combination from C(n,k), you can identify whether the first element of your original set belongs to the subset (which corresponds to X), if you check whether X belongs to either subspace:

• X < C(n-1, k-1) : belongs
• X >= C(n-1, k-1): doesn't belong

Then you can recursively apply the same approach for C(n-1, ...) and so on, until you've found the answer for all n elements of the original set.

Python code to illustrate this approach:

To prove that it's correct, you have to find some sort of invariant. Something that's true during every pass of the loop.

Looking at it, after the very first pass of the inner loop, the largest element of the list will actually be in the first position.

Now in the second pass of the inner loop, i = 1, and the very first comparison is between i = 1 and j = 0. So, the largest element was in position 0, and after this comparison, it will be swapped to position 1.

In general, then it's not hard to see that after each step of the outer loop, the largest element will have moved one to the right. So after the full steps, we know at least the largest element will be in the correct position.

What about all the rest? Let's say the second-largest element sits at position i of the current loop. We know that the largest element sits at position i-1 as per the previous discussion. Counter j starts at 0. So now we're looking for the first A[j] such that it's A[j] > A[i]. Well, the A[i] is the second largest element, so the first time that happens is when j = i-1, at the first largest element. Thus, they're adjacent and get swapped, and are now in the "right" order. Now A[i] again points to the largest element, and hence for the rest of the inner loop no more swaps are performed.

So we can say: Once the outer loop index has moved past the location of the second largest element, the second and first largest elements will be in the right order. They will now slide up together, in every iteration of the outer loop, so we know that at the end of the algorithm both the first and second-largest elements will be in the right position.

What about the third-largest element? Well, we can use the same logic again: Once the outer loop counter i is at the position of the third-largest element, it'll be swapped such that it'll be just below the second largest element (if we have found that one already!) or otherwise just below the first largest element.

Ah. And here we now have our invariant: After k iterations of the outer loop, the k-length sequence of elements, ending at position k-1, will be in sorted order:

After the 1st iteration, the 1-length sequence, at position 0, will be in the correct order. That's trivial.

After the 2nd iteration, we know the largest element is at position 1, so obviously the sequence A[0], A[1] is in the correct order.

Now let's assume we're at step k, so all the elements up to position k-1 will be in order. Now i = k and we iterate over j. What this does is basically find the position at which the new element needs to be slotted into the existing sorted sequence so that it'll be properly sorted. Once that happens, the rest of the elements "bubble one up" until now the largest element sits at position i = k and no further swaps happen.

Thus finally at the end of step N, all the elements up to position N-1 are in the correct order, QED.

I probably don't fully understand the question, because right now the solution is quite trivial. EDIT: yes, I misunderstood after all, the OP does not just want an optimal solution, but wishes to randomly sample from the set of optimal solutions. This answer is not incorrect but it also is an answer to a different question than what OP is interested in.

Simply sort the numbers and greedily construct the subset: