print "X" star patterns in Kotlin - kotlin
can someone help me make star patterns like this using for loop in kotlin? i already try this but i think my code is too long. can someone help me?
x Star Pattern
fun fifthPyramid(){
for(i in 1..13){
if(i==1||i==13){
print("*")
}else
print(" ")
}
println("")
for(i in 1..13){
if(i==2||i==12){
print("*")
}else
print(" ")
}
println("")
for(i in 1..13){
if(i==3||i==11){
print("*")
}else
print(" ")
}
println("")
for(i in 1..13){
if(i==4||i==10){
print("*")
}else
print(" ")
}
println("")
for(i in 1..13){
if(i==5||i==9){
print("*")
}else
print(" ")
}
}
The star pattern consists of exactly N * 2 - 1 rows and columns. So the outer and inner loop will run till towards count = N * 2 - 1
fun main() {
var starCount = 5;
val count = starCount * 2 - 1;
for(i in 1..count){
for(j in 1..count){
if(j==i || (j==count - i + 1))
{
print("*");
}
else
{
print(" ");
}
}
println("")
}
}
Identify the pattern with respect to row and column. You put a * when the row and column are the same, or the row and inverse of the column are the same.
fun printX(size: Int, char: Char) {
repeat(size) { row ->
repeat(size) { col ->
print(if (row == col || row == (size - col - 1)) char else ' ')
}
println()
}
}
fun main() {
printX(7, '*')
}
You can write a fun that takes an Int and a Char as arguments and then use a loop in order to build each line and print it.
Basically like this:
fun printX(heightWidth: Int, symbol: Char) {
// iterate the heigth in order to build up each line (top->down)
for (i in 0 until heightWidth) {
/*
build up an array of Chars (representing a line)
with the desired size (length of a line)
filled up with whitespaces by default
*/
var line = CharArray(heightWidth) { ' ' }
// then replace whitespaces at the desired indexes by the symbol
line[i] = symbol // one from left to right
line[line.size - i - 1] = symbol // and one from right to left
// and print the result
println(line)
}
}
You can throw an Exception in case of negative heightWidth values because they will cause trouble if you don't. And maybe forbid 0, 1 and 2, because although they would produce valid output, that output can hardly be regarded as X. Even the output for 3 and 4 is rather ugly ;-)
However, here's the output of printX(7, '7'):
7 7
7 7
7 7
7
7 7
7 7
7 7
Related
problems with index of array
I'm writing a function that allows you to remove certain numbers from an int arraylist. My code for (i in 1 until 4) { divider = setDivider(i) for(index in 0 until numbers.size){ if(index <= numbers.size){ if (numbers[index] % divider == 0 && !isDone) { numbers.removeAt(index) } }else{ isDone = true } } if(isDone) break } the function to set the divider fun setDivider(divider: Int): Int { when (divider) { 1 -> return 2 2 -> return 3 3 -> return 5 4 -> return 7 } return 8 } I do not know why the ide is giving me the error Index 9 out of bounds for length 9.
Author explained in the comments that the goal is to remove all numbers that are divisible by 2, 3, 5 and 7. It can be achieved much easier by utilizing ready to use functions from stdlib: val dividers = listOf(2, 3, 5, 7) numbers.removeAll { num -> dividers.any { num % it == 0 } } It removes elements that satisfy the provided condition (is divisible) for any of provided dividers. Also, it is often cleaner to not modify a collection in-place, but to create an entirely new collection: val numbers2 = numbers.filterNot { num -> dividers.any { num % it == 0 } }
How to remake the program so that words are passed in function arguments in the KOTLIN programming language?
Need to create a function that implements the attached algorithm, to which all words are passed in the function arguments. For example: f ("dfd" dd "ddd"); My code: fun main() { var s = readLine(); var w = Array(128){0} //To mark characters from a word 1 var g = Array(128){0}//When we encounter a space, we add units from the first array to the corresponding elements of the second, zeroing them in the first. if(s!=null) { for(c in s) { if(c.toInt() > 127 || c.toInt()<0) { println("Input error, try again"); return; } //Checking for space. if(c.toInt() != 32) w[c.toInt()] = 1; else for(k in 0..127) { if(w[k] == 1) { g[k] += 1; w[k] = 0; } } } //For the last word, if there was no space after it. for(k in 0..127) { if(w[k] == 1) { g[k] += 1; w[k] = 0; } } } //Displaying matched characters to the screen for(k in 0..127) { if(g[k]>1) { println(k.toChar()); } } } This program searches for characters that match at least two words in a string Example input: hello world output: lo
There's already utilities for these in Kotlin, I highly recommend you to read the docs before asking these type of questions. The groupingBy should do what you want: readLine()?.let { input -> input.groupingBy { it }.eachCount() .forEach { if (it.value > 1 && it.key != ' ') println(it.key) } }
how to increase the size limit of a mutable list in kotlin?
I was attempting to solve the multiset question (https://codeforces.com/contest/1354/problem/D) on codeforces using Fenwick Tree Data structure. I passed the sample test cases but got the memory limit error after submitting, the testcase is mentioned below. (Basically the testcase is: 1000000 1000000 1.............1 //10^6 times -1...........-1 //10^6 times). I tried similar testcase in my IDE and got the below mentioned error. (Similar to above, the testcase I provided is: 1000000 1 1.............1 //10^6 times -1 ) Exception in thread "main" java.lang.IndexOutOfBoundsException: Index 524289 out of bounds for length 524289 at java.base/jdk.internal.util.Preconditions.outOfBounds(Preconditions.java:64) at java.base/jdk.internal.util.Preconditions.outOfBoundsCheckIndex(Preconditions.java:70) at java.base/jdk.internal.util.Preconditions.checkIndex(Preconditions.java:248) at java.base/java.util.Objects.checkIndex(Objects.java:373) at java.base/java.util.ArrayList.get(ArrayList.java:426) at MultisetKt.main(multiset.kt:47) at MultisetKt.main(multiset.kt) Here is my code: private fun readInt() = readLine()!!.split(" ").map { it.toInt() } fun main() { var (n, q) = readInt() var list = readInt() //modify the list to store it from index 1 var finalList = listOf(0) + list val query = readInt() var bit = MutableList(n+1){0} fun update(i:Int, value:Int) { var index = i while(index < n){ bit.set (index , bit[index] + value) index += (index and -index) } } fun rangefunc(i:Int): Int { var su = 0 var index = i while(index > 0){ su += bit[index] index -= (index and -index) } return su } fun find(x:Int):Int { var l = 1 var r = n var ans = n var mid = 0 while (l <= r) { mid = (l + r) / 2 if (rangefunc(mid) >= x) { ans = mid r = mid - 1 } else { l = mid + 1 } } return ans } for (i in 1..n) { update(finalList[i], 1) } for (j in 0..q - 1) { if (query[j] > 0) { update(query[j], 1) } else { update(find(-query[j]), -1) } } if(rangefunc(n) == 0){ println(0) }else{ println(find(1)) } } I believe this is because the BITlist is not able to store 10^6 elements but not sure. Please let me know what changes should I make in my code also any additional advice on how to deal with such cases in the future. Thank you in advance :)
An ArrayList can store over 2 billion items (2 * 10^9). That is not your issue. ArrayIndexOutOfBoundsException is for trying to access an index of an ArrayList that is less than zero or greater than or equal to its size. In other words, an index that it doesn't yet contain. There's more code there than I have time to debug. But I would start at the line that the stack trace points to and see how it's possible for you to attempt to call bit[index] with an index that equals the size of the ArrayList. To answer your literal question, you can use LinkedList explicitly as your type of MutableList to avoid the size restriction, but it is heavier and it is slower when accessing elements by index.
Kotlin decomposing numbers into powers of 2
Hi I am writing an app in kotlin and need to decompose a number into powers of 2. I have already done this in c#, PHP and swift but kotlin works differently somehow. having researched this I believe it is something to do with the numbers in my code going negative somewhere and that the solution lies in declaring one or more of the variable as "Long" to prevent this from happening but i have not been able to figure out how to do this. here is my code: var salads = StringBuilder() var value = 127 var j=0 while (j < 256) { var mask = 1 shl j if(value != 0 && mask != 0) { salads.append(mask) salads.append(",") } j += 1 } // salads = (salads.dropLast()) // removes the final "," println("Salads = $salads") This shoud output the following: 1,2,4,8,16,32,64 What I actually get is: 1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648,1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072,262144,524288,1048576,2097152,4194304,8388608,16777216,33554432,67108864,134217728,268435456,536870912,1073741824,-2147483648, Any ideas?
This works for the one input that you specified, at the very least: fun powersOfTwo(value :Long): String { val result = ArrayList<String>() var i = 0 var lastMask = 0 while (lastMask < value) { val mask = 1 shl i if (value != 0.toLong() && mask < value) { result.add(mask.toString()) } lastMask = mask i += 1 } return result.joinToString(",") } Ran it in a unit test: #Test fun addition_isCorrect() { val result = powersOfTwo(127) assertEquals("1,2,4,8,16,32,64", result) } Test passed.
You can get a list of all powers of two that fit in Int and test each of them for whether the value contains it with the infix function and: val value = 126 val powersOfTwo = (0 until Int.SIZE_BITS).map { n -> 1 shl n } println(powersOfTwo.filter { p -> value and p != 0}.joinToString(",")) // prints: 2,4,8,16,32,64 See the entire code in Kotlin playground: https://pl.kotl.in/f4CZtmCyI
Hi I finally managed to get this working properly: fun decomposeByTwo(value :Int): String { val result = ArrayList<String>() var value = value var j = 0 while (j < 256) { var mask = 1 shl j if ((value and mask) != 0) { value -= mask result.add(mask.toString()) } j += 1 } return result.toString() } I hope this helps someone trying to get a handle on bitwise options!
Somehow you want to do the "bitwise AND" of "value" and "mask" to determine if the j-th bit of "value" is set. I think you just forgot that test in your kotlin implementation.
How do I return from an anonymous recursive sub in perl6?
This does what I'd expect. fib(13) returns 233. sub fib(Int $a --> Int) { return 0 if $a == 0; return 1 if $a == 1; return fib($a -1) + fib($a -2); } my $square = -> $x { $x * 2 }; # this works with no return value my #list = <1 2 3 4 5 6 7 8 9>.map( $square ); # returns [2 4 6 8 10 12 14 16 18] I tried implementing fib() using an anonymous sub my $fib = -> Int $x --> Int { return 0 if $x == 0; return 1 if $x == 1; return $fib($x - 1) + $fib($x - 2); } $fib(13) I get the following error when running that with explicit returns. Attempt to return outside of any Routine in block at test.p6 line 39 So I got rid of the return values. my $fib = -> Int $x --> Int { 0 if $x == 0; 1 if $x == 1; $fib($x - 1) + $fib($x - 2); } say $fib(13); This last version never returns. Is there a way to write an anonymous recursive function without return values?
According to the documentation : Blocks that aren't of type Routine (which is a subclass of Block) are transparent to return. sub f() { say <a b c>.map: { return 42 }; # ^^^^^^ exits &f, not just the block } The last statement is the implicit return value of the block So you can try: my $fib = -> Int $x --> Int { if ( $x == 0 ) { 0; # <-- Implicit return value } elsif ( $x == 1 ) { 1; # <-- Implicit return value } else { $fib($x - 1) + $fib($x - 2); # <-- Implicit return value } }
Three more options: sub You can write anonymous routines by using sub without a name: my $fib = sub (Int $x --> Int) { return 0 if $x == 0; return 1 if $x == 1; return $fib($x - 1) + $fib($x - 2); } say $fib(13); # 233 See #HåkonHægland's answer for why this (deliberately) doesn't work with non-routine blocks. leave The design anticipated your question: my $fib = -> Int $x --> Int { leave 0 if $x == 0; leave 1 if $x == 1; leave $fib($x - 1) + $fib($x - 2); } compiles. Hopefully you can guess that what it does -- or rather is supposed to do -- is exactly what you wanted to do. Unfortunately, if you follow the above with: say $fib(13); You get a run-time error "leave not yet implemented". My guess is that this'll get implemented some time in the next few years and the "Attempt to return outside of any Routine" error message will then mention leave. But implementing it has very low priority because it's easy to write sub as above, or write code as #HåkonHægland did, or use a case/switch statement construct as follows, and that's plenty good enough for now. case/switch (when/default) You can specify the parameter as $_ instead of $x and then you're all set to use constructs that refer to the topic: my $fib = -> Int $_ --> Int { when 0 { 0 } when 1 { 1 } $fib($_ - 1) + $fib($_ - 2) } say $fib(13); # 233 See when.
Blocks don't need to declare the return type. You can still return whatever you want, though. The problem is not in using return, it's in the declaration of the Int. use v6; my $fib = -> Int $x { if $x == 0 { 0; } elsif $x == 1 { 1; } else { $fib($x - 1) + $fib($x - 2); } } say $fib(13) ; The problem is that the return value needs to be the last executed. In the way you have done it, if it finds 0 or 1 it keeps running, getting to the last statement, when it will start all over again. Alternatively, you can use given instead of the cascaded ifs. As long as whatever it returns is the last issued, it's OK.