# swift-algorithm-club | Algorithms and data structures in Swift | Learning library

## kandi X-RAY | swift-algorithm-club Summary

## kandi X-RAY | swift-algorithm-club Summary

Here you'll find implementations of popular algorithms and data structures in everyone's favorite new language Swift, with detailed explanations of how they work.

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## swift-algorithm-club Key Features

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## Community Discussions

Trending Discussions on swift-algorithm-club

QUESTION

I've been studying up on graphs using the Adjacency List implementation, and I am reading that adding an edge is an O(1) operation.

This makes sense if you are just tacking an edge on to the Vertex's Linked List of edges, but I don't understand how this could be the case so long as you care about removing the old edge if **one already exists**. Finding that edge would take O(V) time.

If you don't do this, and you add an edge that already exists, you would have duplicate entries for that edge, which means they could have different weights, etc.

Can anyone explain what I seem to be missing? Thanks!

...ANSWER

Answered 2018-Aug-21 at 06:22You're right at your complecxity analysis. Find if edge already exist is truly O(V). But notice that adding this edge even if existed is still O(1).

You need to remember that having 2 edges with the same source an destination are **valid** input to graph - even with different weights (maybe not even but because).

That way adding edge to adjacency-list-graph is O(1)

QUESTION

Based on this Ray Wenderlich article I'm able to create a binary tree data structure like that:

...ANSWER

Answered 2019-Apr-08 at 00:04In terms of the basics of how to draw a line, you:

- Create
`UIBezierPath`

; - Move to the starting point with
`move(to:)`

; - Adding line to the end point with
`addLine(to:)`

;

You can then render that path in your UI by either:

- Create
`CAShapeLayer`

, specifying its`strokeWidth`

,`strokeColor`

, and`fillColor`

; set its`path`

, and then add that shape layer as a sublayer of the view’s`layer`

; or - Create
`UIView`

subclass and in its`draw(_:)`

method you can call`setStroke`

of the desired`UIColor`

, set the`lineWidth`

of the`UIBezierPath`

, and then`stroke()`

the`UIBezierPath`

.

Generally, I’d use the `CAShapeLayer`

approach, where I basically configure the shape layer, but let the OS render that shape layer for me.

That having been said, I’d probably take this a step further, and wrap the line drawing in its own `UIView`

subclass. The thought process is that not only are high-level views generally composed of `UIView`

objects, but it also opens the door for all sorts of advanced UX (e.g. you might want to detect a tap on a node and do something).

Anyway, I’d wrap that “connector line” drawing code in a `UIView`

subclass, like so:

QUESTION

Going through Swift algorithms including the brute force string search i.e.: https://github.com/raywenderlich/swift-algorithm-club/tree/master/Brute-Force%20String%20Search

The output is said to be 7

Now actually the output is (String.Index?) - and although I can subscript with that I want to actually see the value - 7.

The code is

...ANSWER

Answered 2019-Feb-26 at 03:06Try this out:

QUESTION

ANSWER

Answered 2018-Apr-17 at 18:46My solution is a little bit more wordy :) where I replaced printTree with my own two methods

QUESTION

I have code like this:

...ANSWER

Answered 2018-Mar-11 at 04:26Map your generic type `T`

to `String`

and then apply `joined()`

function.

`joined`

is only for elements of type `String`

and here compiler do not know the type of `T`

.

QUESTION

I have a series of coordinates that I'm wanting to find the linear regression for, so I can then find the bearing of the line. I'm using the Linear Regression algorithm from Swift Algorithm Club on this set of coordinates:

...ANSWER

Answered 2017-Sep-17 at 08:55I confirm that your bearing function works correctly. However, the regression value you get for latitude is 51.45437262420372 - which is where the regression line intersects with longitude = 0.0. I think you need to use both the intersection point and slope of the regression line to calculate fitted lat,lon values (within your map region), and then find the bearing using those values.

LATER ADDITION: An important issue here is that longitude coordinates are not linear in relation to distance - the distance covered by a degree of longitude varies with latitude. I think the simplest solution is just to convert your array of lat,lon values to lat,normalisedLon values where normalisedLon = lon * cosine(lat).

When you do this the angle of your path from your first coordinate to your last coordinate = 3.055 degrees (just east of north which appears to be correct).

Once you've done this then the slope of your regression line will in fact represent the direction you are seeking i.e. heading = atan(slope).

However, your regression code just doesn't seem to work for me. It should give a result very close to 3 degrees also, but it is way off.

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