String str= -PT31121936-1-0069902679870--BLUECH
I want divide the above string by useing string Tokenize
output like this:
amount=" ";
txnNo = PT31121936;
SeqNo = 1;
AccNo = 0069902679870;
Cldflag=" ";
FundOption= BLUECH;
Solution in Java using String split, it would be better than String tokenizer.
There are two solutions
1) This approach is assuming the input string will be always in a specific order.
2) This approach is more dynamic, where we can accommodate change in the order of the input string and also in the number of parameters. My preference would be the second approach.
public class StringSplitExample {
public static void main(String[] args) {
// This solution is based on the order of the input
// is Amount-Txn No-Seq No-Acc No-Cld Flag-Fund Option
String str= "-PT31121936-1-0069902679870--BLUECH";
String[] tokens = str.split("-");
System.out.println("Amount :: "+tokens[0]);
System.out.println("Txn No :: "+tokens[1]);
System.out.println("Seq No :: "+tokens[2]);
System.out.println("Acc No :: "+tokens[3]);
System.out.println("Cld Flag :: "+tokens[4]);
System.out.println("Fund Option :: "+tokens[5]);
// End of First Solution
// The below solution can take any order of input, but we need to provide the order of input
String[] tokensOrder = {"Txn No", "Amount", "Seq No", "Cld Flag", "Acc No", "Fund Option"};
String inputString = "PT31121936--1--0069902679870-BLUECH";
String[] newTokens = inputString.split("-");
// Check whether both arrays are having equal count - To avoid index out of bounds exception
if(newTokens.length == tokensOrder.length) {
for(int i=0; i<tokensOrder.length; i++) {
System.out.println(tokensOrder[i]+" :: "+newTokens[i]);
}
}
}
}
Reference: String Tokenizer vs String split
Scanner vs. StringTokenizer vs. String.Split
Related
import java.util.Scanner;
class Palindrome_string
{
public static void main()
{
System.out.println("\f");
Scanner sc = new Scanner(System.in);
System.out.println("Enter a string");
String a = sc.nextLine();
int b = a.length();
String rev = "";
for (int i = b - 1; i >= 0; i--)
{
char c = a.charAt(i);
rev = rev + c;
}
System.out.println("Original word "+a);
System.out.println("Reversed word "+rev);
a = a.toLowerCase();
rev = rev.toLowerCase();
if (a == rev)
{
System.out.println("It is a palindrome");
}
else
{
System.out.println("It is not a palindrome");
}
sc.close();
}
}
The program compiles properly. Still, when running the program, the message which tells if it is a palindrome prints incorrectly. What changes do I make? Here is a picture of the output. Even though the word 'level' (which is a palindrome) has been inputted, it shows that it isn't a palindrome. What changes should I make? output pic
You should not use == to compare two strings because it compares the reference of the string, i.e. whether they are the same object or not.
Use .equals() instead. It tests for value equality. So in your case:
if (a.equals(rev))
{
System.out.println("It is a palindrome");
}
Also try not to use single-letter variable names except for index variables when iterating over a list etc. It's bad practice.
public class JavaIntern{
public static void main(String []args){
String str = "JavaIntern"; //hardcoded, but not the problem
char[] s = str.toCharArray();
String result = new String (s,0,1); //this is where the dilemma begins
System.out.println(result);
String result1 = new String (s,0,2);
System.out.println(result1);
String result2 = new String (s,0,3);
System.out.println(result2);
String result3 = new String (s,0,4);
System.out.println(result3);
String result4 = new String (s,0,5);
System.out.println(result4);
String result5 = new String (s,0,6);
System.out.println(result5);
String result6 = new String (s,0,7);
System.out.println(result6);
String result7 = new String (s,0,8);
System.out.println(result7);
String result8 = new String (s,0,9);
System.out.println(result8);
String result9 = new String (s,0,10);
System.out.println(result9); //and this is where it ends... how can I get rid of this?
}
}
//but still get this:
J
Ja
Jav
Java
JavaI
JavaIn
JavaInt
JavaInte
JavaInter
JavaIntern
I guess you want to improve the code and also don't depend on the length of the string.
What about something like this?
public class JavaIntern{
public static void main(String []args){
String str = "JavaIntern"; //hardcoded, but not the problem
String substring = "";
for (char ch: str.toCharArray()) {
substring += ch;
System.out.println(substring);
}
}
}
This will also print:
J
Ja
Jav
Java
JavaI
JavaIn
JavaInt
JavaInte
JavaInter
JavaIntern
The loop gets one character of the string at a time and concatenates it to the substring before printing it.
Im assuming you want to be able to print out one letter more each time.
To do this we use a for loop, and this way it is fairly simple.
public class MainClass {
public static void main(String[] args) {
String str = "JavaIntern";
for (int i = 1; i <= str.length(); i++) {
System.out.println(str.substring(0, i));
}
}
}
We set i to 0 in the loop, keep iterating while i less than or equal to the length of the string, and each time we iterate, add one to i.
We use the substring method to split the string from the first letter, to i.
I am new to Dart programming language and anyone help me find the best string concatenation methods available in Dart.
I only found the plus (+) operator to concatenate strings like other programming languages.
There are 3 ways to concatenate strings
String a = 'a';
String b = 'b';
var c1 = a + b; // + operator
var c2 = '$a$b'; // string interpolation
var c3 = 'a' 'b'; // string literals separated only by whitespace are concatenated automatically
var c4 = 'abcdefgh abcdefgh abcdefgh abcdefgh'
'abcdefgh abcdefgh abcdefgh abcdefgh';
Usually string interpolation is preferred over the + operator.
There is also StringBuffer for more complex and performant string building.
If you need looping for concatenation, I have this :
var list = ['satu','dua','tiga'];
var kontan = StringBuffer();
list.forEach((item){
kontan.writeln(item);
});
konten = kontan.toString();
Suppose you have a Person class like.
class Person {
String name;
int age;
Person({String name, int age}) {
this.name = name;
this.age = age;
}
}
And you want to print the description of person.
var person = Person(name: 'Yogendra', age: 29);
Here you can concatenate string like this
var personInfoString = '${person.name} is ${person.age} years old.';
print(personInfoString);
Easiest way
String get fullname {
var list = [firstName, lastName];
list.removeWhere((v) => v == null);
return list.join(" ");
}
The answer by Günter covers the majority of the cases of how you would concatenate two strings in Dart.
If you have an Iterable of Strings it will be easier to use writeAll as this gives you the option to specify an optional separator
final list = <String>['first','second','third'];
final sb = StringBuffer();
sb.writeAll(list, ', ');
print(sb.toString());
This will return
'first, second, third'
Let's think we have two strings
String a = 'Hello';
String b = 'World';
String output;
Now we want to concat this two strings
output = a + b;
print(output);
Hello World
I would like to convert the string containing abc to a list of characters and a hashset of characters. How can I do that in Java ?
List<Character> charList = new ArrayList<Character>("abc".toCharArray());
In Java8 you can use streams I suppose.
List of Character objects:
List<Character> chars = str.chars()
.mapToObj(e->(char)e).collect(Collectors.toList());
And set could be obtained in a similar way:
Set<Character> charsSet = str.chars()
.mapToObj(e->(char)e).collect(Collectors.toSet());
You will have to either use a loop, or create a collection wrapper like Arrays.asList which works on primitive char arrays (or directly on strings).
List<Character> list = new ArrayList<Character>();
Set<Character> unique = new HashSet<Character>();
for(char c : "abc".toCharArray()) {
list.add(c);
unique.add(c);
}
Here is an Arrays.asList like wrapper for strings:
public List<Character> asList(final String string) {
return new AbstractList<Character>() {
public int size() { return string.length(); }
public Character get(int index) { return string.charAt(index); }
};
}
This one is an immutable list, though. If you want a mutable list, use this with a char[]:
public List<Character> asList(final char[] string) {
return new AbstractList<Character>() {
public int size() { return string.length; }
public Character get(int index) { return string[index]; }
public Character set(int index, Character newVal) {
char old = string[index];
string[index] = newVal;
return old;
}
};
}
Analogous to this you can implement this for the other primitive types.
Note that using this normally is not recommended, since for every access you
would do a boxing and unboxing operation.
The Guava library contains similar List wrapper methods for several primitive array classes, like Chars.asList, and a wrapper for String in Lists.charactersOf(String).
The lack of a good way to convert between a primitive array and a collection of its corresponding wrapper type is solved by some third party libraries. Guava, a very common one, has a convenience method to do the conversion:
List<Character> characterList = Chars.asList("abc".toCharArray());
Set<Character> characterSet = new HashSet<Character>(characterList);
Use a Java 8 Stream.
myString.chars().mapToObj(i -> (char) i).collect(Collectors.toList());
Breakdown:
myString
.chars() // Convert to an IntStream
.mapToObj(i -> (char) i) // Convert int to char, which gets boxed to Character
.collect(Collectors.toList()); // Collect in a List<Character>
(I have absolutely no idea why String#chars() returns an IntStream.)
The most straightforward way is to use a for loop to add elements to a new List:
String abc = "abc";
List<Character> charList = new ArrayList<Character>();
for (char c : abc.toCharArray()) {
charList.add(c);
}
Similarly, for a Set:
String abc = "abc";
Set<Character> charSet = new HashSet<Character>();
for (char c : abc.toCharArray()) {
charSet.add(c);
}
List<String> result = Arrays.asList("abc".split(""));
Create an empty list of Character and then make a loop to get every character from the array and put them in the list one by one.
List<Character> characterList = new ArrayList<Character>();
char arrayChar[] = abc.toCharArray();
for (char aChar : arrayChar)
{
characterList.add(aChar); // autoboxing
}
You can do this without boxing if you use Eclipse Collections:
CharAdapter abc = Strings.asChars("abc");
CharList list = abc.toList();
CharSet set = abc.toSet();
CharBag bag = abc.toBag();
Because CharAdapter is an ImmutableCharList, calling collect on it will return an ImmutableList.
ImmutableList<Character> immutableList = abc.collect(Character::valueOf);
If you want to return a boxed List, Set or Bag of Character, the following will work:
LazyIterable<Character> lazyIterable = abc.asLazy().collect(Character::valueOf);
List<Character> list = lazyIterable.toList();
Set<Character> set = lazyIterable.toSet();
Bag<Character> set = lazyIterable.toBag();
Note: I am a committer for Eclipse Collections.
IntStream can be used to access each character and add them to the list.
String str = "abc";
List<Character> charList = new ArrayList<>();
IntStream.range(0,str.length()).forEach(i -> charList.add(str.charAt(i)));
Using Java 8 - Stream Funtion:
Converting A String into Character List:
ArrayList<Character> characterList = givenStringVariable
.chars()
.mapToObj(c-> (char)c)
.collect(collectors.toList());
Converting A Character List into String:
String givenStringVariable = characterList
.stream()
.map(String::valueOf)
.collect(Collectors.joining())
To get a list of Characters / Strings -
List<String> stringsOfCharacters = string.chars().
mapToObj(i -> (char)i).
map(c -> c.toString()).
collect(Collectors.toList());
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I saw this in an interview question ,
Given a sorting order string, you are asked to sort the input string based on the given sorting order string.
for example if the sorting order string is dfbcae
and the Input string is abcdeeabc
the output should be dbbccaaee.
any ideas on how to do this , in an efficient way ?
The Counting Sort option is pretty cool, and fast when the string to be sorted is long compared to the sort order string.
create an array where each index corresponds to a letter in the alphabet, this is the count array
for each letter in the sort target, increment the index in the count array which corresponds to that letter
for each letter in the sort order string
add that letter to the end of the output string a number of times equal to it's count in the count array
Algorithmic complexity is O(n) where n is the length of the string to be sorted. As the Wikipedia article explains we're able to beat the lower bound on standard comparison based sorting because this isn't a comparison based sort.
Here's some pseudocode.
char[26] countArray;
foreach(char c in sortTarget)
{
countArray[c - 'a']++;
}
int head = 0;
foreach(char c in sortOrder)
{
while(countArray[c - 'a'] > 0)
{
sortTarget[head] = c;
head++;
countArray[c - 'a']--;
}
}
Note: this implementation requires that both strings contain only lowercase characters.
Here's a nice easy to understand algorithm that has decent algorithmic complexity.
For each character in the sort order string
scan string to be sorted, starting at first non-ordered character (you can keep track of this character with an index or pointer)
when you find an occurrence of the specified character, swap it with the first non-ordered character
increment the index for the first non-ordered character
This is O(n*m), where n is the length of the string to be sorted and m is the length of the sort order string. We're able to beat the lower bound on comparison based sorting because this algorithm doesn't really use comparisons. Like Counting Sort it relies on the fact that you have a predefined finite external ordering set.
Here's some psuedocode:
int head = 0;
foreach(char c in sortOrder)
{
for(int i = head; i < sortTarget.length; i++)
{
if(sortTarget[i] == c)
{
// swap i with head
char temp = sortTarget[head];
sortTarget[head] = sortTarget[i];
sortTarget[i] = temp;
head++;
}
}
}
In Python, you can just create an index and use that in a comparison expression:
order = 'dfbcae'
input = 'abcdeeabc'
index = dict([ (y,x) for (x,y) in enumerate(order) ])
output = sorted(input, cmp=lambda x,y: index[x] - index[y])
print 'input=',''.join(input)
print 'output=',''.join(output)
gives this output:
input= abcdeeabc
output= dbbccaaee
Use binary search to find all the "split points" between different letters, then use the length of each segment directly. This will be asymptotically faster then naive counting sort, but will be harder to implement:
Use an array of size 26*2 to store the begin and end of each letter;
Inspect the middle element, see if it is different from the element left to it. If so, then this is the begin for the middle element and end for the element before it;
Throw away the segment with identical begin and end (if there are any), recursively apply this algorithm.
Since there are at most 25 "split"s, you won't have to do the search for more than 25 segemnts, and for each segment it is O(logn). Since this is constant * O(logn), the algorithm is O(nlogn).
And of course, just use counting sort will be easier to implement:
Use an array of size 26 to record the number of different letters;
Scan the input string;
Output the string in the given sorting order.
This is O(n), n being the length of the string.
Interview questions are generally about thought process and don't usually care too much about language features, but I couldn't resist posting a VB.Net 4.0 version anyway.
"Efficient" can mean two different things. The first is "what's the fastest way to make a computer execute a task" and the second is "what's the fastest that we can get a task done". They might sound the same but the first can mean micro-optimizations like int vs short, running timers to compare execution times and spending a week tweaking every millisecond out of an algorithm. The second definition is about how much human time would it take to create the code that does the task (hopefully in a reasonable amount of time). If code A runs 20 times faster than code B but code B took 1/20th of the time to write, depending on the granularity of the timer (1ms vs 20ms, 1 week vs 20 weeks), each version could be considered "efficient".
Dim input = "abcdeeabc"
Dim sort = "dfbcae"
Dim SortChars = sort.ToList()
Dim output = New String((From c In input.ToList() Select c Order By SortChars.IndexOf(c)).ToArray())
Trace.WriteLine(output)
Here is my solution to the question
import java.util.*;
import java.io.*;
class SortString
{
public static void main(String arg[])throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
// System.out.println("Enter 1st String :");
// System.out.println("Enter 1st String :");
// String s1=br.readLine();
// System.out.println("Enter 2nd String :");
// String s2=br.readLine();
String s1="tracctor";
String s2="car";
String com="";
String uncom="";
for(int i=0;i<s2.length();i++)
{
if(s1.contains(""+s2.charAt(i)))
{
com=com+s2.charAt(i);
}
}
System.out.println("Com :"+com);
for(int i=0;i<s1.length();i++)
if(!com.contains(""+s1.charAt(i)))
uncom=uncom+s1.charAt(i);
System.out.println("Uncom "+uncom);
System.out.println("Combined "+(com+uncom));
HashMap<String,Integer> h1=new HashMap<String,Integer>();
for(int i=0;i<s1.length();i++)
{
String m=""+s1.charAt(i);
if(h1.containsKey(m))
{
int val=(int)h1.get(m);
val=val+1;
h1.put(m,val);
}
else
{
h1.put(m,new Integer(1));
}
}
StringBuilder x=new StringBuilder();
for(int i=0;i<com.length();i++)
{
if(h1.containsKey(""+com.charAt(i)))
{
int count=(int)h1.get(""+com.charAt(i));
while(count!=0)
{x.append(""+com.charAt(i));count--;}
}
}
x.append(uncom);
System.out.println("Sort "+x);
}
}
Here is my version which is O(n) in time. Instead of unordered_map, I could have just used a char array of constant size. i.,e. char char_count[256] (and done ++char_count[ch - 'a'] ) assuming the input strings has all ASCII small characters.
string SortOrder(const string& input, const string& sort_order) {
unordered_map<char, int> char_count;
for (auto ch : input) {
++char_count[ch];
}
string res = "";
for (auto ch : sort_order) {
unordered_map<char, int>::iterator it = char_count.find(ch);
if (it != char_count.end()) {
string s(it->second, it->first);
res += s;
}
}
return res;
}
private static String sort(String target, String reference) {
final Map<Character, Integer> referencesMap = new HashMap<Character, Integer>();
for (int i = 0; i < reference.length(); i++) {
char key = reference.charAt(i);
if (!referencesMap.containsKey(key)) {
referencesMap.put(key, i);
}
}
List<Character> chars = new ArrayList<Character>(target.length());
for (int i = 0; i < target.length(); i++) {
chars.add(target.charAt(i));
}
Collections.sort(chars, new Comparator<Character>() {
#Override
public int compare(Character o1, Character o2) {
return referencesMap.get(o1).compareTo(referencesMap.get(o2));
}
});
StringBuilder sb = new StringBuilder();
for (Character c : chars) {
sb.append(c);
}
return sb.toString();
}
In C# I would just use the IComparer Interface and leave it to Array.Sort
void Main()
{
// we defin the IComparer class to define Sort Order
var sortOrder = new SortOrder("dfbcae");
var testOrder = "abcdeeabc".ToCharArray();
// sort the array using Array.Sort
Array.Sort(testOrder, sortOrder);
Console.WriteLine(testOrder.ToString());
}
public class SortOrder : IComparer
{
string sortOrder;
public SortOrder(string sortOrder)
{
this.sortOrder = sortOrder;
}
public int Compare(object obj1, object obj2)
{
var obj1Index = sortOrder.IndexOf((char)obj1);
var obj2Index = sortOrder.IndexOf((char)obj2);
if(obj1Index == -1 || obj2Index == -1)
{
throw new Exception("character not found");
}
if(obj1Index > obj2Index)
{
return 1;
}
else if (obj1Index == obj2Index)
{
return 0;
}
else
{
return -1;
}
}
}