spiral | Light-weight Web Audio/MIDI API Library | Audio Utils library
kandi X-RAY | spiral Summary
kandi X-RAY | spiral Summary
Spiral is a JavaScript library for Web Audio API. It is a light-weight version of WAAX core library. Its goal is to extend Web Audio API while minimizing the abstraction and the layer.
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Currently covering the most popular Java, JavaScript and Python libraries. See a Sample of spiral
spiral Key Features
spiral Examples and Code Snippets
public static List spiralOrder(int[][] matrix) {
int row = matrix.length, col = matrix[0].length;
int left = 0, right = 0, top = 0, down = 0;
int r = 0, c = 0;
List sol = new ArrayList<>();
while (sol.siz
public List spiralOrder(int[][] matrix) {
List result = new ArrayList();
if(matrix == null || matrix.length == 0) {
return result;
}
int rowStart = 0;
int rowEnd = matrix.length - 1;
def spiral_matrix(matrix):
n = len(matrix)
if n == 0:
return []
m = len(matrix[0])
if m == 0:
return []
total = n * m
res = []
n -= 1
xDir, yDir = 1, 1
x, y = 0, -1
while len(res) < total
Community Discussions
Trending Discussions on spiral
QUESTION
given a spiral of numbers in which the numbers are arranged in the form of a triangle, i need to write a function that takes a number and returns the coordinates of this number
...ANSWER
Answered 2021-May-27 at 10:26TL;DR: Working code at the bottom
First consider how many numbers you have in each triangle:
- The first contains 3 numbers in each edge
- The second contains 6 numbers in each edge
- The third contains 9 numbers in each edge
So there is our pattern. Triangle i
contains 9i
numbers and the before we do anything else we need to divide our number by 9
and find the triangle root of the result (and round it down):
Once you figure out the right triangle, there are three things left to do:
- You need to find the starting point
This is trivial as the "last point" of triangle i
will always be (2i, i)
.
- You need to find the right edge
You already know that your triangle has i
-long edges so by taking the sum of your remainders (from the original divmod
and from rooting) and dividing it by 3 you can find the right edge.
- You need to find the right point on this edge
This bit is trivial - you have 3 types of edges, horizontal, vertical and diagonal. Depending on the type you have to apply your "final residual value" to the "origin of the edge":
(-r, -r)
for the diagonal(0, r)
for the vertical(r, 0)
for the horizontal
Relative to the max point of the previous triangle to get to the right edge you just have to apply a cumulative sum of these transpositions:
(-r, -r)
for the diagonal(-i, -i+r)
for the vertical(-i+r, 0)
for the horizontal
Putting it all together
QUESTION
"terminal.integrated.shellArgs.windows": [
"-ExecutionPolicy", "ByPass", "-NoExit", "-Command", "& 'C:/Users/Marko/anaconda3/shell/condabin/conda-hook.ps1' ; conda activate 'C:/Users/Marko/anaconda3'"
]
...ANSWER
Answered 2021-May-27 at 09:06Check out the answer here. Just running conda init
will set things up set the Anaconda env gets initialized on Powershell's startup.
QUESTION
I have a database with all the students of a school, the year of each class and the class number. As you need to go to school for 5 years in my country, each student has 5 lines with his name in this database, but with each time different class so the line is not the same.
I want SQL to select classmates in my database. All classes are composed with different classmates but you can be two year in a row with the same classmate.
I want to know if a student has been in the same class with student 1 and student 2 but not at the same time. If a student has been with student 1 in year 1 and with student 2 in year 3 it is ok, like this is a common student. In the picture the correct answer is Lucien Lake.
I tried this code, but it does not work because it only tells me all the students for student Paul Biloux.
...ANSWER
Answered 2021-May-16 at 18:15First you need to list out all classmate pairs by year, then join that result to itself for the same student in different years. The most recent SQLLite supports common table expressions which makes this query look pretty clean:
QUESTION
I am trying to making a python autogenerated Email app but there is a problem when running the code the traceback error shows up but I did write the code as my mentor write it down. This is the code that I used:
...ANSWER
Answered 2021-May-18 at 03:10Try and set the encoding to UTF-8
For example:
file = open(filename, encoding="utf8")
For reference check this post:
UnicodeDecodeError: 'charmap' codec can't decode byte X in position Y: character maps to
QUESTION
Here is a sample of my database with 3 colums : class, year and student_name.
I search the "distance" in terms of common student between two given student. (Like in the famous game "6 degrees of Bacon")
The goal is to calculate the distance between Momo Thary (2018) and Paul Biloux (2020). And the answer is 2 students : Momo Thary was in the same class than Jack Spiral. Jack Spiral was in the same class than Lucien Lake. And Lucien Lake was in the same class than Paul Biloux.
So we can "link" Momo Thary and Paul Biloux with 2 students.
I am using SQLite online. With the following code, it is possible to know the common player of Jack Spiral and Paul Biloux :
...ANSWER
Answered 2021-May-17 at 14:01You can use a recursive query :
QUESTION
I am trying to make a grid out of angular spirals. The spiral itself is composed of single lines within a for loop. When I duplicate and shift (translate) the origin of the spiral along one axis (x OR y) it works. But shifting along both (x AND y), in order to make it a grid, it does not work out without decomposing the spiral.
I would really appreciate if anyone could help me with my coding puzzle. By the way I'm very open-minded for any tips and help improving my code writing skills. There surely is a lot of redundancy and long-winded expressions in there... This is my code so far:
...ANSWER
Answered 2021-May-07 at 21:08If you're trying to make a grid of spirals it looks like you just need to use a pair of for loops where you currently have for (let j = 0; j < 5; j++) {
. Pretty much any time you want to create a grid you're going to want a pair of nested for loops.
QUESTION
I have a Navbar with logo and links spaced out appropriately, in the hero section I use an image with full width and height but if leaves a white space between the nav and hero section. I have searched for how to address this but cannot seem to figure it out.
How can I remove the space between the Nav and next section?
Example Image: https://ibb.co/7YcTg4p
*Solved - After adding overflow: auto;
inside the #container-bg {}
class the white space collapsed and now the nav follows the next section with any space issues.
ANSWER
Answered 2021-May-05 at 11:08I think I did not understand the question but it may help
QUESTION
i'm super new to XSL programming and i have the following xml;
...ANSWER
Answered 2021-Apr-30 at 12:43The output you show can be produced quite easily by:
XSLT 1.0
QUESTION
I cant get FB_MBReadinputs to work in Twincat, when Factory IO is sending/receiving Input and Holding registers.
First off here's my currently working snip of handling Modbus from Factory IO:
...ANSWER
Answered 2021-Mar-29 at 12:44Just a few things I noticed:
- MBReadRegs will be writing to your array GAB_FactoryIO_RegsIN but you have this variable configured in the input space %I*. Is there a reason for that?
- In your second example you are reading in 16 words but the destination variable is only 12 words (ARRAY [0..5] OF DINT). That could possibly be the cause of the ADS error 1794, "invalid index group". With MBReadRegs, nQuantity refers to 16-bit words whereas MBReadInputs counts bits.
- You are triggering the bExecute bit on every scan. Generally you should trigger it once and wait for the bBusy bit to go false before triggering again.
QUESTION
I have made a simple python program that creates randomly generated spirals using the turtle module. I am still pretty new at python and i was wondering how i could restart the program after the turtle is pretty far away from the center and the spiral is done. Here is my code:
...ANSWER
Answered 2021-Apr-12 at 17:43You could do it by using a global flag variable and installing a turtle timer that calls a function that periodically checks to see whether a spiral has finished being drawn yet by looking at the variable — and if so, resets things and draws another one. In the code below the function that does that is named check_status()
. I also added a short pause before the after each one is finished to allow it to remain visible long enough to be admired. ;¬)
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