I want to write a code that regards strings with characters in different order as equal as long as same characters are placed in the strings. For example, suppose $a = "ksv", whenever somebody input string value "svk" or "kvs", I want to write a code that these strings are equivalent to $a. Here is an example,
#ans=("ksv", "kvs", "svk", "vsk",......);
if (#input[1] ~~ #ans) {
return 'EXACT_ANS';
}
#input[1] is what user will put the string. At first, I listed all of the different types of order as an array (just like the example) so that if one of elements in the array matches with #input[1], then I return it as a correct answer. However, this is quite a long and tedious work if I have a string with much longer length. Please give me any advice on this. Thank you^^
You want something of the form
if (normalize_string($input) eq normalize_string('ksv')) {
...
}
where normalize_string is a sub that returns the same string for all equivalent inputs, and returns different strings for inputs that aren't equivalent.
The exact definition of normalize_string will vary based on what you consider equivalent.
If you want to ignore duplicate characters (abbc is equivalent to abc):
sub normalize_string {
my %h;
++$h{$_} for split //, $_[0];
return join '', sort keys %h;
}
If the number of instances of each character is pertinent (abbc isn't equivalent to abc):
sub normalize_string {
return join '', sort split //, $_[0];
}
Of course, you can inline the normalized form when the parameter is a constant.
if (normalize_string($input) eq 'ksv') {
...
}
Related
I have an array which outputs the following:
charges = [5.00, 26.00, 8.00, 4.00, 4.00, -8.00, 54.00, 52.48]
When I try to perform a sum using this:
charges.sum()
It gives me:
5.0026.008.004.004.00-8.0054.0052.48
I am assuming I need to convert it from a string to a float so I did:
Float.valueOf((String) charges.sum())
and it gives me an error which states 'multiple points'.
My question is how do I add all of these figures up?
If your list is actually of strings, you can quickly do the conversion with sum()'s closure form.
charges.sum { it.toBigDecimal() }
It is unclear what your list has in it, it seems like the entries in your list are actually strings (since sum concatenates them), or something that prints a string value (has a toString method that returns a string showing the value so you think it is numeric even though it isn’t) . They are definitely not numeric, so convert each of them to a numeric type before summing:
charges.collect { new BigDecimal(it.toString()) }.sum()
(What your code was doing was concatenating the string values together, then converting that to a numeric type.)
You must delete the cast (String)
Float.valueOf(charges.sum())
So the idea here is that I'm taking a .csv into a string and each value needs to be stored into a variable. I am unsure how to properly parse a string to do this.
My idea is a function that looks like
final char delim = ',';
int nextItem(String data, int startFrom) {
if (data.charAt(startFrom) != delim) {
return data.charAt(startFrom)
} else {
return nextItem(data, startFrom + 1);
}
}
so if I passed it something like
nextItem("45,621,9", 0);
it would return 45
and if I passed it
nextItem("45,621,9", 3);
it would return 621
I'm not sure if I have that setup properly to be recursive, but I could also use a For loop I suppose, only real stipulation is I can't use the Substring method.
Please don't use recursion for a matter that can be easily done iteratively. Recursion is expensive in terms of stack and calling frames: A very long string could produce a StackOverflowError.
I suggest you take a look to standard method indexOf of java.lang.String:
A good alternative is Regular Expressions.
You can seperate the words considering comma ',' as delimeter
Code
String[] nextItem(String data) {
String[] words=data.split(",");
return words;
}
This will return an array of strings that is the words in your input string. Then you can use the array in anyway you need.
Hope it helps ;)
Processing comes with a split() function that does exactly what you're describing.
From the reference:
String men = "Chernenko,Andropov,Brezhnev";
String[] list = split(men, ',');
// list[0] is now "Chernenko", list[1] is "Andropov"...
Behind the scenes it's using the String#split() function like H. Sodi's answer, but you should just use this function instead of defining your own.
I am trying to solve this problem on hackerrank:
So the problem is:
Jack and Daniel are friends. Both of them like letters, especially upper-case ones.
They are cutting upper-case letters from newspapers, and each one of them has their collection of letters stored in separate stacks.
One beautiful day, Morgan visited Jack and Daniel. He saw their collections. Morgan wondered what is the lexicographically minimal string, made of that two collections. He can take a letter from a collection when it is on the top of the stack.
Also, Morgan wants to use all the letters in the boys' collections.
This is my attempt in Perl:
#!/usr/bin/perl
use strict;
use warnings;
chomp(my $n=<>);
while($n>0){
chomp(my $string1=<>);
chomp(my $string2=<>);
lexi($string1,$string2);
$n--;
}
sub lexi{
my($str1,$str2)=#_;
my #str1=split(//,$str1);
my #str2=split(//,$str2);
my $final_string="";
while(#str2 && #str1){
my $st2=$str2[0];
my $st1=$str1[0];
if($st1 le $st2){
$final_string.=$st1;
shift #str1;
}
else{
$final_string.=$st2;
shift #str2;
}
}
if(#str1){
$final_string=$final_string.join('',#str1);
}
else{
$final_string=$final_string.join('',#str2);
}
print $final_string,"\n";
}
Sample Input:
2
JACK
DANIEL
ABACABA
ABACABA
The first line contains the number of test cases, T.
Every next two lines have such format: the first line contains string A, and the second line contains string B.
Sample Output:
DAJACKNIEL
AABABACABACABA
But for Sample test-case it is giving right results while it is giving wrong results for other test-cases. One case for which it gives an incorrect result is
1
AABAC
AACAB
It outputs AAAABACCAB instead of AAAABACABC.
I don't know what is wrong with the algorithm and why it is failing with other test cases?
Update:
As per #squeamishossifrage comments If I add
($str1,$str2)=sort{$a cmp $b}($str1,$str2);
The results become same irrespective of user-inputs but still the test-case fails.
The problem is in your handling of the equal characters. Take the following example:
ACBA
BCAB
When faced with two identical characters (C in my example), you naïvely chose the one from the first string, but that's not always correct. You need to look ahead to break ties. You may even need to look many characters ahead. In this case, next character after C of the second string is lower than the next character of the first string, so you should take the C from the second string first.
By leaving the strings as strings, a simple string comparison will compare as many characters as needed to determine which character to consume.
sub lexi {
my ($str1, $str2) = #_;
utf8::downgrade($str1); # Makes sure length() will be fast
utf8::downgrade($str2); # since we only have ASCII letters.
my $final_string = "";
while (length($str2) && length($str1)) {
$final_string .= substr($str1 le $str2 ? $str1 : $str2, 0, 1, '');
}
$final_string .= $str1;
$final_string .= $str2;
print $final_string, "\n";
}
Too little rep to comment thus the answer:
What you need to do is to look ahead if the two characters match. You currently do a simple le match and in the case of
ZABB
ZAAA
You'll get ZABBZAA since the first match Z will be le Z. So what you need to do (a naive solution which most likely won't be very effective) is to keep looking as long as the strings/chars match so:
Z eq Z
ZA eq ZA
ZAB gt ZAA
and at that point will you know that the second string is the one you want to pop from for the first character.
Edit
You updated with sorting the strings, but like I wrote you still need to look ahead. The sorting will solve the two above strings but will fail with these two:
ZABAZA
ZAAAZB
ZAAAZBZABAZA
Because here the correct answer is ZAAAZABAZAZB and you can't find that will simply comparing character per character
Suppose I have a string "That question is on the minds of every one.".
I want to compare each word in string with a set of word I.e. (to , is ,on , of) and if those word occurs I want to append some string on the existing string.
Eg.
to = append "Hi";
Is = append "Hello";
And so on.
To be more specific I have used StringTokenizer to get the each word and compared thru if else statement. However we can use Switch also but it is available in Jdk 1.
7.
I don't know if this is what you mean, but:
You could use String.split() to separate the words from your string like
String[] words = myString.split(" ");
and then, for each word, compare it with the given set
for(String s : words)
{
switch(s)
{
case("to"):
[...]
}
}
Or you could just use the String.contains() method without even splitting your string, but I don't know if that's what you wanted.
Use a HashMap<String,String> variable to store your set of words and the replacement words you want. Then split your string with split(), loop through the resulting String[] and for each String in the String[], check whether the HashMap containsKey() that String. Build your output/resulting String in the loop - if the word is contained in the HashMap, replace it with the value of the corresponding key in the HashMap, otherwise use the String you are currently on from the String[].
I am using the play framework 2.2.1 and I have a question concerning the manipulation of Strings within view templates. Unfortunately I am not very familiar with the Scala programming language nor its APIs. The strings are contained in a List which is passed from the controller to the view and then I use a loop to process each string before they are added to the html. I would like to know how to do the following: trim, toLowerCase and remove spaces. As an example, if I have "My string ", I would like to produce "mystring". More specifically I would actually like to produce "myString", however I'm sure I can figure that out if someone points me in the right direction. Thanks.
UPDATE:
Fiaz provided a great solution, building on his answer and just for interest sake I came up with the following solution using recursion. This example is of course making many assumptions about the input provided.
#formatName(name: String) = #{
def inner(list: List[String], first: Boolean): String = {
if (!list.tail.isEmpty && first) list.head + inner(list.tail, false)
else if (!list.tail.isEmpty && !first) list.head.capitalize + inner(list.tail, false)
else if (list.tail.isEmpty && !first) list.head.capitalize
else list.head
}
if (!name.trim.isEmpty) inner(name.split(' ').map(_.toLowerCase).toList, true)
else ""
}
If you want to know how to do just the trimming, lower-casing and joining without spaces, try this perhaps?
// Given that s is your string
s.split(" ").map(_.toLowerCase).mkString
That splits a string into an array strings, splitting is done on one or more spaces so that gives you trimmed strings. You then map each element in the array with the function (x => x.toLowerCase) (for which the shorthand is (_.toLowerCase)) and then join the Array back into a single string using the mkString method that collections have.
So let's say you want to capitalize the first letter of the each of the space-split bits:
Scala provides a capitalize method on Strings, so you could use that:
s.split(" ").map(_.toLowerCase.capitalize).mkString
See http://www.scala-lang.org/api/current/scala/collection/immutable/StringOps.html
One suggestion as to how you can get the exact output (your example 'myString') you describe:
(s.split(" ").toList match {
case fst::rest => fst.toLowerCase :: rest.map(_.toLowerCase.capitalize)
case Nil => Nil }
).mkString
There is example of using the string manipulation below:
#stringFormat(value: String) = #{
value.replace("'", "\\'")
}
#optionStringFormat(description: Option[String]) = #{
if (description.isDefined) {
description.get.replace("'", "\\'").replace("\n", "").replace("\r", "")
} else {
""
}
}
#for(photo <- photos) {
<div id="photo" class="random" onclick="fadeInPhoto(#photo.id, '#photo.filename', '#stringFormat(photo.title)', '#optionStringFormat(photo.description)', '#byTags');">
This example obtained from https://github.com/joakim-ribier/play2-scala-gallery