swap | Development repository for the swap cookbook | Infrastructure Automation library

yarn and npm already do this job, why not use them?

Tag the dev versions when you release them

Rather than using two different function, one for adding card image and one for card content, try combining both of them.. here use the code for your reference.

According to the Go memory model:

https://golang.org/ref/mem

There are no guarantees that one goroutine will see the operations performed by another goroutine unless there is an explicit synchronization between the two using channels, mutex. etc.

In your example: the fact that a goroutines sees done=true does not imply it will see a set. This is only guaranteed if there is explicit synchronization between the goroutines.

The sync.Once probably offers such synchronization, so that's why you have not observed this behavior. There is still a memory race, and on a different platform with a different implementation of sync.Once, things may change.

Let's call an element that is equal to its position a "wrong" element.

Let 𝑛 be the number of wrong elements. The goal is to reduce 𝑛 to zero.

This problem would have been more complex if duplicate values were allowed, but given they are distinct, we can observe the following:

• A swap is only useful if it reduces 𝑛. If 𝑛 is not decreased by it, there is no way to take benefit from that swap in future swaps.
• If a swap reduces 𝑛 by 2, it may still not be an optimal swap. For example, if in [1,2,3,4] we first swap (2,3), then we will need 2 more swaps to bring 𝑛 to zero, while this can be solved using just two swaps. We can avoid this by starting with the first pair (or last pair,...) when dealing with a group of adjacent wrong elements.

• If the array has only one element and it is wrong -- i.e. the input is [1] -- then there is no solution possible.
• Iterate through the array from left to right.
• When arriving at a wrong element, perform a single swap with the next element if there is one, otherwise swap with the previous one.
• Return the number of swaps made.

We can actually improve on this, and only count the swaps, but not actually perform them. In that case we must skip the next element when we count a swap of the current element with the next. This is to avoid we would count another swap for it when it is also bad, as the current swap would also resolve that. On the other hand, this swap never makes the next one bad, so when counting this swap we can safely skip checking the next element.

Here is a JavaScript snippet that returns the result for several inputs:

We can't send mail on_behalf_of with a guest user account because a guest user doesn't have O365 Exchange Online license in this tenant.

Although maybe the guest user has O365 Exchange Online license in its own tenant, it is not allowed to send mail from this tenant.

It is expected.

What can I do to bypass this?

If you want the guest user to send the mail from this tenant, it's impossible.

But I think sending mail from its own tenant is not what you want and it will require you to create app registration in that tenant or use multi-tenant app. You need admin permission of that tenant to do that. So it's impossible neither.

The fix to this seemed to be to wrap the datepicker in a $(document).ready(function() { }) type function.

So the full datepicker function from above becomes:

You can store them with a GADT, like this:

I am not sure that it is possible for multiple lines, but for one, this is the way in linux:

It would have half of previous time complexity i guess.

O(0.5 * n^2) is still O(n^2). A good qsort() is expected O(n* ln(n)).

Is this efficient enough or should I stick with selection sort and qsort.

Tough to beat decades of many programmers experience.

Keep in mind qsort() does not have to use the quick sort algorithm. A good qsort() may use a combination of algorithms.