minimum | Minimum , android app launcher | Runtime Evironment library

Items will be aligned if both QGroupBoxes have same count of children and each row have at least one child with Expanding vertical policy. Instead of spacer use QWidget

I removed unrelated widgets and reduces number of rows to 4 for demonstration purposes (less xml).

There are two problems which are causing the counter not to be incremented.

The first is that the CSS:

Starting from version 0.5 of Rocket, you are not expected to use rocket_contrib, because this one was split into features which are either already in the core crate or moved to separate crates. The notes from this revision (see also issue 1659) provide a few more details:

This follows the completed graduation of stable contrib features into core, removing 'rocket_contrib' in its entirety in favor of two new crates. These crates are versioned independently of Rocket's core libraries, allowing upgrades to dependencies without consideration for versions in core libraries.

'rocket_dyn_templates' replaces the contrib 'templates' features. While largely a 1-to-1 copy, it makes the following changes:

• the 'tera_templates' feature is now 'tera'
• the 'handlebars_templates' feature is now 'handlebars'
• fails to compile if neither 'tera' nor 'handlebars' is enabled

'rocket_sync_db_pools' replaces the contrib 'database' features. It makes no changes to the replaced features except that the database attribute is properly documented at the crate root.

In short, you will need to migrate your code away from rocket_contrib. Better guidelines may become available once v0.5 is definitely released, but until then, you may look for the features once available in rocket_contrib in the core documentation and respective Cargo feature list.

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:

For improve performance first sorting values per groups (by both columns), then get differencies, replace first misisng value by first value (minimal) and get mean:

I can see the recurse relation as T(n) = T(n-1) + T(n-2) + T(n-3) + T(n-4) + ... + T(0), since the loop is from l to h (ignore the if condition for now). So for an interval [l,h] every value in that interval will be called in the worst case that is minJumps(l+1, h), minJumps(l+2, h) ... minJumps(h, h) and it can be noticed that the above recurse relation holds here.

Now, solving the relation, we can write it as T(n) = T(n-1) + T(n-1) as T(n-1) = T(n-2) + T(n-3) + T(n-4) + ... + T(0). Hence T(n) = 2 * T(n-1) which boils down to O(2^n).

The time complexity of the mentioned algorithm should be O(2^n).

I would do something like this:

here is how you can do it: I assumed LastUpdated column is the column from tableA which indicate date of

Normally, I would do something with scipy.ndimage.morphology.distance_transform_edt but this gives the assume that isotropic sizes of the voxel:

It does no such thing! You are looking for the sampling= parameter. From the latest version of the docs:

Spacing of elements along each dimension. If a sequence, must be of length equal to the input rank; if a single number, this is used for all axes. If not specified, a grid spacing of unity is implied.

The wording "sampling" or "spacing" is probably a bit mysterious if you think of pixels as little squares/cubes, and that is probably why you missed it. In most situations, it is better to think of pixels as point samples on a grid, with fixed spacing between samples. I recommend Alvy Ray's a pixel is not a little square for a better understanding of this terminology.