I have a rudimentary Flow layout which will break line as soon as the items exceed the maximum width.
I want to put a divider between each element, except when it is the first or last item of a line. I made a custom layout for this.
(Simplified, without the logic for where to add the separator; just to illustrate I cannot place a Placeable multiple times:)
#Composable
private fun FlowRowWithDivider(
divider: #Composable () -> Unit,
modifier: Modifier = Modifier,
content: #Composable () -> Unit,
) {
Layout(
{
divider()
content()
},
modifier,
) { measurables, constr ->
val placeables = measurables.map { it.measure(constr) }
val separator = placeables[0]
val sepBaseline = separator[FirstBaseline]
layout(constr.maxWidth, constr.maxHeight) {
var yPos = 0
var xPos = 0
var maxY = 0
placeables.sublist(start = 1).forEach { placeable ->
if (xPos + placeable.width > constr.maxWidth) {
xPos = 0
yPos += maxY
maxY = 0
}
placeable.placeRelative(x = xPos, y = yPos)
xPos += placeable.width
val charBaseline = placeable[FirstBaseline]
separator.place(xPos + placeable.width, if (sepBaseline == AlignmentLine.Unspecified) charBaseline - sepBaseline else 0)
if (maxY < placeable.height) {
maxY = placeable.height
}
}
}
}
}
This does not work, because placeables can be placed only once. How can I generate one placeable for each divider that I need? I was looking into accessing currentComposer to generate new nodes, but I did not understand it.
Is there a way to 'stamp' the divider Composable multiple times?
Related
Given a string, find the maximum deviation among all substrings. The maximum deviation is defined as the difference between the maximum frequency of a character and the minimum frequency of a character.
For example, in abcaba, a has a frequency of 3; b has a frequency of 2; c has a frequency of 1. so a has the maximum frequency, which is 3, whereas c has a minimum frequency of 1. Therefore the deviation of this string is 3 - 1 = 2. And we also need to find all other deviations for each of the substrings for abacaba, the maximum among them is the answer.
I couldn't think of a better way rather than the obvious brute force approach. Thanks in advance!
For finding all substrings you have to consider O(n2). See this post for more details. You can just optimize it by stop point where substring lengths be smaller than current maximum deviation.
maxDeviation = 0;
n = strlen(str);
for i = 0 to n
{
if(n-i < maxDeviation) break; //this is new stop point to improve
sub1 = substring(str,i,n);
sub2 = substring(str,0,n-i); // you can use if(i!=0) to avoid duplication of first sentence
a = findMaxDeviation(sub1); // This is O(n)
b = findMaxDeviation(sub2); // This is O(n)
maxDeviation = max(a,b);
}
print maxDeviation
Pay attention to this line if(n-i < maxDeviation) break; because you cannot find a deviation more than maxDeviation in a string with length of smaller than maxDeviation.
public static int getDev(Map<String, Integer> devEntries){
List<Integer> entries = devEntries.entrySet().stream()
.map(x->x.getValue())
.collect(Collectors.toList());
Comparator<Integer> collect = Comparator.naturalOrder();
Collections.sort(entries,collect.reversed());
return entries.get(0) - entries.get( entries.size()-1);
}
public static int getMaxFreqDeviation(String s, Set<Integer> deviations ) {
for (int x=0;x<s.length();x++) {
for (int g=x;g<s.length()+1;g++){
String su =s.substring(x,g);
Map<String, Integer> map = Arrays.asList(su.split(""))
.stream()
.collect(Collectors.groupingBy(v->v,Collectors.summingInt(v->1)));
if (map.entrySet().size()==1){
deviations.add(abs(0));
}else {
int devcount = getDev(map);
deviations.add(abs(devcount));
}
}
}
return deviations.stream().collect(Collectors.toList()).get(deviations.size()-1);
}
public static void main(String[] args){
String se = "abcaba";
Set<Integer> deviations = new TreeSet<>();
int ans = getMaxFreqDeviation(se,deviations);
System.out.println(ans);
}
}
I faced a similar question in a test and I used c#, although I failed during the challenge but picked it up to solve the next day. I came about something like the below.
var holdDict = new Dictionary<char, int>();
var sArray = s.ToCharArray();
var currentCharCount = 1;
//Add the first element
holdDict.Add(sArray[0],1);
for (int i = 1; i < s.Length-1; i++)
{
if (sArray[i] == sArray[i - 1])
{
currentCharCount += 1;
}
else
{
currentCharCount = 1;
}
holdDict.TryGetValue(sArray[i], out var keyValue);
if (keyValue < currentCharCount) holdDict[sArray[i]] = currentCharCount;
}
var myQueue = new PriorityQueue<string, int>();
foreach (var rec in holdDict)
{
myQueue.Enqueue($"{rec.Key}#{rec.Value}", rec.Value);
}
int highest = 0, lowest = 0, queueCount=myQueue.Count;
while (myQueue.Count > 0)
{
int currentValue = int.Parse(myQueue.Peek().Split('#')[1]);
if (myQueue.Count == queueCount) lowest = currentValue;
highest = currentValue;
myQueue.Dequeue();
}
return highest - lowest;
O(n) algo (26*26*N)
import string
def maxSubarray(s, ch1, ch2):
"""Find the largest sum of any contiguous subarray."""
"""From https://en.wikipedia.org/wiki/Maximum_subarray_problem"""
best_sum = 0
current_sum = 0
for x in s:
if x == ch1:
x = 1
elif x == ch2:
x = -1
else:
x = 0
current_sum = max(0, current_sum + x)
best_sum = max(best_sum, current_sum)
return best_sum
def findMaxDiv(s):
'''Algo from https://discuss.codechef.com/t/help-coding-strings/99427/4'''
maxDiv = 0
for ch1 in string.ascii_lowercase:
for ch2 in string.ascii_lowercase:
if ch1 == ch2:
continue
curDiv = maxSubarray(s, ch1, ch2)
if curDiv > maxDiv:
maxDiv = curDiv
return maxDiv
I want to make my dot program turn around when they reach edge
so basically i just simply calculate
x = width/2+cos(a)*20;
y = height/2+sin(a)*20;
it's make circular movement. so i want to make this turn around by checking the edge. i also already make sure that y reach the if condition using println command
class particles {
float x, y, a, r, cosx, siny;
particles() {
x = width/2; y = height/2; a = 0; r = 20;
}
void display() {
ellipse(x, y, 20, 20);
}
void explode() {
a = a + 0.1;
cosx = cos(a)*r;
siny = sin(a)*r;
x = x + cosx;
y = y + siny;
}
void edge() {
if (x>width||x<0) cosx*=-1;
if (y>height||y<0) siny*=-1;
}
}
//setup() and draw() function
particles part;
void setup(){
size (600,400);
part = new particles();
}
void draw(){
background(40);
part.display();
part.explode();
part.edge();
}
they just ignore the if condition
There is no problem with your check, the problem is with the fact that presumably the very next time through draw() you ignore what you did in response to the check by resetting the values of cosx and siny.
I recommend creating two new variables, dx and dy ("d" for "direction") which will always be either +1 and -1 and change these variables in response to your edge check. Here is a minimal example:
float a,x,y,cosx,siny;
float dx,dy;
void setup(){
size(400,400);
background(0);
stroke(255);
noFill();
x = width/2;
y = height/2;
dx = 1;
dy = 1;
a = 0;
}
void draw(){
ellipse(x,y,10,10);
cosx = dx*20*cos(a);
siny = dy*20*sin(a);
a += 0.1;
x += cosx;
y += siny;
if (x > width || x < 0)
dx = -1*dx;
if (y > height || y < 0)
dy = -1*dy;
}
When you run this code you will observe the circles bouncing off the edges:
I have been trying to implement parallel merge sort in Scala. But with 8 cores, using .sorted is still about twice as fast.
edit:
I rewrote most of the code to minimize object creation. Now it runs about as fast as the .sorted
Input file with 1.2M integers:
1.333580 seconds (my implementation)
1.439293 seconds (.sorted)
How should I parallelize this?
New implementation
object Mergesort extends App
{
//=====================================================================================================================
// UTILITY
implicit object comp extends Ordering[Any] {
def compare(a: Any, b: Any) = {
(a, b) match {
case (a: Int, b: Int) => a compare b
case (a: String, b: String) => a compare b
case _ => 0
}
}
}
//=====================================================================================================================
// MERGESORT
val THRESHOLD = 30
def inssort[A](a: Array[A], left: Int, right: Int): Array[A] = {
for (i <- (left+1) until right) {
var j = i
val item = a(j)
while (j > left && comp.lt(item,a(j-1))) {
a(j) = a(j-1)
j -= 1
}
a(j) = item
}
a
}
def mergesort_merge[A](a: Array[A], temp: Array[A], left: Int, right: Int, mid: Int) : Array[A] = {
var i = left
var j = right
while (i < mid) { temp(i) = a(i); i+=1; }
while (j > mid) { temp(i) = a(j-1); i+=1; j-=1; }
i = left
j = right-1
var k = left
while (k < right) {
if (comp.lt(temp(i), temp(j))) { a(k) = temp(i); i+=1; k+=1; }
else { a(k) = temp(j); j-=1; k+=1; }
}
a
}
def mergesort_split[A](a: Array[A], temp: Array[A], left: Int, right: Int): Array[A] = {
if (right-left == 1) a
if ((right-left) > THRESHOLD) {
val mid = (left+right)/2
mergesort_split(a, temp, left, mid)
mergesort_split(a, temp, mid, right)
mergesort_merge(a, temp, left, right, mid)
}
else
inssort(a, left, right)
}
def mergesort[A: ClassTag](a: Array[A]): Array[A] = {
val temp = new Array[A](a.size)
mergesort_split(a, temp, 0, a.size)
}
Previous implementation
Input file with 1.2M integers:
4.269937 seconds (my implementation)
1.831767 seconds (.sorted)
What sort of tricks there are to make it faster and cleaner?
object Mergesort extends App
{
//=====================================================================================================================
// UTILITY
val StartNano = System.nanoTime
def dbg(msg: String) = println("%05d DBG ".format(((System.nanoTime - StartNano)/1e6).toInt) + msg)
def time[T](work: =>T) = {
val start = System.nanoTime
val res = work
println("%f seconds".format((System.nanoTime - start)/1e9))
res
}
implicit object comp extends Ordering[Any] {
def compare(a: Any, b: Any) = {
(a, b) match {
case (a: Int, b: Int) => a compare b
case (a: String, b: String) => a compare b
case _ => 0
}
}
}
//=====================================================================================================================
// MERGESORT
def merge[A](left: List[A], right: List[A]): Stream[A] = (left, right) match {
case (x :: xs, y :: ys) if comp.lteq(x, y) => x #:: merge(xs, right)
case (x :: xs, y :: ys) => y #:: merge(left, ys)
case _ => if (left.isEmpty) right.toStream else left.toStream
}
def sort[A](input: List[A], length: Int): List[A] = {
if (length < 100) return input.sortWith(comp.lt)
input match {
case Nil | List(_) => input
case _ =>
val middle = length / 2
val (left, right) = input splitAt middle
merge(sort(left, middle), sort(right, middle + length%2)).toList
}
}
def msort[A](input: List[A]): List[A] = sort(input, input.length)
//=====================================================================================================================
// PARALLELIZATION
//val cores = Runtime.getRuntime.availableProcessors
//dbg("Detected %d cores.".format(cores))
//lazy implicit val ec = ExecutionContext.fromExecutorService(Executors.newFixedThreadPool(cores))
def futuremerge[A](fa: Future[List[A]], fb: Future[List[A]])(implicit order: Ordering[A], ec: ExecutionContext) =
{
for {
a <- fa
b <- fb
} yield merge(a, b).toList
}
def parallel_msort[A](input: List[A], length: Int)(implicit order: Ordering[A]): Future[List[A]] = {
val middle = length / 2
val (left, right) = input splitAt middle
if(length > 500) {
val fl = parallel_msort(left, middle)
val fr = parallel_msort(right, middle + length%2)
futuremerge(fl, fr)
}
else {
Future(msort(input))
}
}
//=====================================================================================================================
// MAIN
val results = time({
val src = Source.fromFile("in.txt").getLines
val header = src.next.split(" ").toVector
val lines = if (header(0) == "i") src.map(_.toInt).toList else src.toList
val f = parallel_msort(lines, lines.length)
Await.result(f, concurrent.duration.Duration.Inf)
})
println("Sorted as comparison...")
val sorted_src = Source.fromFile(input_folder+"in.txt").getLines
sorted_src.next
time(sorted_src.toList.sorted)
val writer = new PrintWriter("out.txt", "UTF-8")
try writer.print(results.mkString("\n"))
finally writer.close
}
My answer is probably going to be a bit long, but i hope that it will be useful for both you and me.
So, first question is: "how scala is doing sorting for a List?" Let's have a look at the code from scala repo!
def sorted[B >: A](implicit ord: Ordering[B]): Repr = {
val len = this.length
val b = newBuilder
if (len == 1) b ++= this
else if (len > 1) {
b.sizeHint(len)
val arr = new Array[AnyRef](len) // Previously used ArraySeq for more compact but slower code
var i = 0
for (x <- this) {
arr(i) = x.asInstanceOf[AnyRef]
i += 1
}
java.util.Arrays.sort(arr, ord.asInstanceOf[Ordering[Object]])
i = 0
while (i < arr.length) {
b += arr(i).asInstanceOf[A]
i += 1
}
}
b.result()
}
So what the hell is going on here? Long story short: with java. Everything else is just size justification and casting. Basically this is the line which defines it:
java.util.Arrays.sort(arr, ord.asInstanceOf[Ordering[Object]])
Let's go one level deeper into JDK sources:
public static <T> void sort(T[] a, Comparator<? super T> c) {
if (c == null) {
sort(a);
} else {
if (LegacyMergeSort.userRequested)
legacyMergeSort(a, c);
else
TimSort.sort(a, 0, a.length, c, null, 0, 0);
}
}
legacyMergeSort is nothing but single threaded implementation of merge sort algorithm.
The next question is: "what is TimSort.sort and when do we use it?"
To my best knowledge default value for this property is false, which leads us to TimSort.sort algorithm. Description can be found here. Why is it better? Less comparisons that in merge sort according to comments in JDK sources.
Moreover you should be aware that it is all single threaded, so no parallelization here.
Third question, "your code":
You create too many objects. When it comes to performance, mutation (sadly) is your friend.
Premature optimization is the root of all evil -- Donald Knuth. Before making any optimizations (like parallelism), try to implement single threaded version and compare the results.
Use something like JMH to test performance of your code.
You should not probably use Stream class if you want to have the best performance as it does additional caching.
I intentionally did not give you answer like "super-fast merge sort in scala can be found here", but just some tips for you to apply to your code and coding practices.
Hope it will help you.
I need a fast method to find all indices of a search term that might occur in a string. I tried this 'brute force' String extension method:
// Note: makes use of ExSwift
extension String
{
var length: Int { return count(self) }
func indicesOf(searchTerm:String) -> [Int] {
var indices = [Int]()
for i in 0 ..< self.length {
let segment = self[i ... (i + searchTerm.length - 1)]
if (segment == searchTerm) {
indices.append(i)
}
}
return indices;
}
}
... But it's ridiculously slow, especially the shorter the search term is. What would be a better method to find all indices fast?
As Martin said you can implement some of the well known fastest algorithms in String Matching, The Knuth–Morris–Pratt string searching algorithm (or KMP algorithm) searches for occurrences of a "word" W within a main "text string" S.
The algorithm has complexity O(n), where n is the length of S and the O is big-O notation.
extension String {
// Build pi function of prefixes
private func build_pi(str: String) -> [Int] {
var n = count(str)
var pi = Array(count: n + 1, repeatedValue: 0)
var k = -1
pi[0] = -1
for (var i = 0; i < n; ++i) {
while (k >= 0 && str[k] != str[i]) {
k = pi[k]
}
pi[i + 1] = ++k
}
return pi
}
// Knuth-Morris Pratt algorithm
func searchPattern(pattern: String) -> [Int] {
var matches = [Int]()
var n = count(self)
var m = count(pattern)
var k = 0
var pi = build_pi(pattern)
for var i = 0; i < n; ++i {
while (k >= 0 && (k == m || pattern[k] != self[i])) {
k = pi[k]
}
if ++k == m {
matches.append(i - m + 1)
}
}
return matches
}
subscript (i: Int) -> Character {
return self[advance(self.startIndex, i)]
}
}
Then you can use it in the following way:
var string = "apurba mandal loves ayoshi loves"
var pattern = "loves"
println(string.searchPattern(pattern))
An the output should be :
[14, 27]
That belong to the start index of the pattern occurrences inside the the string. I hope this help you.
EDIT:
As Martin said in his comment you need to avoid the use of the advance function to index an String by an Int because it's O(position to index).
One possible solution is to convert the String to an array of Character and then access to the indexes is O(1).
Then the extension can be changed to this one :
extension String {
// Build pi function of prefixes
private func build_pi(str: [Character]) -> [Int] {
var n = count(str)
var pi = Array(count: n + 1, repeatedValue: 0)
var k = -1
pi[0] = -1
for (var i = 0; i < n; ++i) {
while (k >= 0 && str[k] != str[i]) {
k = pi[k]
}
pi[i + 1] = ++k
}
return pi
}
// Knuth-Morris Pratt algorithm
func searchPattern(pattern: String) -> [Int] {
// Convert to Character array to index in O(1)
var patt = Array(pattern)
var S = Array(self)
var matches = [Int]()
var n = count(self)
var m = count(pattern)
var k = 0
var pi = build_pi(patt)
for var i = 0; i < n; ++i {
while (k >= 0 && (k == m || patt[k] != S[i])) {
k = pi[k]
}
if ++k == m {
matches.append(i - m + 1)
}
}
return matches
}
}
Instead of checking for the search term at each position of the string
you could use rangeOfString() to find the next occurrence (hoping
that rangeOfString() uses more advanced algorithms):
extension String {
func indicesOf(searchTerm:String) -> [Int] {
var indices = [Int]()
var pos = self.startIndex
while let range = self.rangeOfString(searchTerm, range: pos ..< self.endIndex) {
indices.append(distance(self.startIndex, range.startIndex))
pos = range.startIndex.successor()
}
return indices
}
}
Generally, it depends on the size of the input string and the size
of the search string which algorithm is "the fastest". You'll find
an overview with links to various algorithms in
String searching algorithm.
Update for Swift 3:
extension String {
func indices(of searchTerm:String) -> [Int] {
var indices = [Int]()
var pos = self.startIndex
while let range = range(of: searchTerm, range: pos ..< self.endIndex) {
indices.append(distance(from: startIndex, to: range.lowerBound))
pos = index(after: range.lowerBound)
}
return indices
}
}
Using NSRegularExpression in Swift 4, you can do it like this. NSRegularExpression has been around forever and is probably a better choice than rolling your own algorithm for most cases.
let text = "The quieter you become, the more you can hear."
let searchTerm = "you"
let regex = try! NSRegularExpression(pattern: searchTerm, options: [])
let range: NSRange = NSRange(text.startIndex ..< text.endIndex, in: text)
let matches: [NSTextCheckingResult] = regex.matches(in: text, options: [], range: range)
let ranges: [NSRange] = matches.map { $0.range }
let indices: [Int] = ranges.map { $0.location }
let swiftRanges = ranges.map { Range($0, in: text) }
let swiftIndices: [String.Index] = swiftRanges.flatMap { $0?.lowerBound }
For context: I am going to analyze the breathing movement of parents during kangaroo mother care and I wish to respect their privacy by not recording them, but only the movement of stickers I placed on their chest and stomach.
So far, I'm able to track 2 colours based on webcam input through the code below. However, I would like to record only the tracked colours instead of the webcam feed as to preserve the privacy of the parent.
Does anybody know how to add a background colour, whilst still being able to track colour?
import processing.video.*;
Capture video;
final int TOLERANCE = 20;
float XRc = 0;// XY coordinate of the center of the first target
float YRc = 0;
float XRh = 0;// XY coordinate of the center of the second target
float YRh = 0;
int ii=0; //Mouse click counter
color trackColor; //The first color is the center of the robot
color trackColor2; //The second color is the head of the robot
void setup() {
size(640,480);
video = new Capture(this,640,480);
video.start();
trackColor = color(255,0,0);
trackColor2 = color(255,0,0);
smooth();
}
void draw() {
background(0);
if (video.available()) {
video.read();
}
video.loadPixels();
image(video,0,0);
float r2 = red(trackColor);
float g2 = green(trackColor);
float b2 = blue(trackColor);
float r3 = red(trackColor2);
float g3 = green(trackColor2);
float b3 = blue(trackColor2);
int somme_x = 0, somme_y = 0;
int compteur = 0;
int somme_x2 = 0, somme_y2 = 0;
int compteur2 = 0;
for(int x = 0; x < video.width; x++) {
for(int y = 0; y < video.height; y++) {
int currentLoc = x + y*video.width;
color currentColor = video.pixels[currentLoc];
float r1 = red(currentColor);
float g1 = green(currentColor);
float b1 = blue(currentColor);
if(dist(r1,g1,b1,r2,g2,b2) < TOLERANCE) {
somme_x += x;
somme_y += y;
compteur++;
}
else if(compteur > 0) {
XRc = somme_x / compteur;
YRc = somme_y / compteur;
}
if(dist(r1,g1,b1,r3,g3,b3) < TOLERANCE) {
somme_x2 += x;
somme_y2 += y;
compteur2++;
}
else if(compteur2 > 0) {
XRh = somme_x2 / compteur2;
YRh = somme_y2 / compteur2;
}
}
}
if(XRc != 0 || YRc != 0) { // Draw a circle at the first target
fill(trackColor);
strokeWeight(0.05);
stroke(0);
ellipse(XRc,YRc,20,20);
}
if(XRh != 0 || YRh != 0) {// Draw a circle at the second target
fill(trackColor2);
strokeWeight(0.05);
stroke(0);
ellipse(XRh,YRh,20,20);
}
}
void mousePressed() {
if (mousePressed && (mouseButton == RIGHT)) { // Save color where the mouse is clicked in trackColor variable
if(ii==0){
if (mouseY>480){mouseY=0;mouseX=0;}
int loc = mouseX + mouseY*video.width;
trackColor = video.pixels[loc];
ii=1;
}
else if(ii==1){
if (mouseY>480){mouseY=0;mouseX=0;}
int loc2 = mouseX + mouseY*video.width;
trackColor2 = video.pixels[loc2];
ii=2;
}
}
}
Try adding the background(0); right before you draw the first circle. It should cover the video and you can draw the circles on top of it.
Regards
Jose