Check if string contains any part of substring - java

I have
String S = "Eng - Computer, Eng - Software.."
User inputs:
String I = "Engineering."
I would like this to return true because S contains "Eng" part of a substring of I.
How can I do this?
S.trim.toLowercase.contains(....)
Does not properly work because of the "part" of substring.

Is S a standard string that checks if a substring which enters a user is part of S?
If so, you can use the "contains" method:
if (I.contains("*/string that you want to check*/")
return true;

It depends on what you mean. Are you looking if any of the words in S are contained in I? If that's the case, I would recommend splitting S into an array of Strings for each word, and then checking if any of the words are a substring of I. Take a look at this function.
public boolean checkSubstring(String S, String I){
String[] words = S.split(" ");
for(int i=0; i<words.length; i++){
if(I.contains(words[i])
return true; //We found a word contained in I!
}
return false; //None of the words were contained in I
}

Try the code below:
You need to tokenize S and then ask if I contains any of those tokens.
String S = "Eng - Computer, Eng - Software..";
String I = "Engineering.";
//neither S or I contain each other
System.out.println(S.trim().toLowerCase().contains(I));
System.out.println(I.trim().toLowerCase().contains(S));
String[] tokens = S.split(" ");
//but parts of S are contained in I...
for(String token : tokens) {
if(I.contains(token)) {
System.out.println("found : " +token);
}
}

Related

Replace words in parenthesis

I have a question about replacing words. I have some strings, each of which looks like this:
String string = "today is a (happy) day, I would like to (explore) more about Java."
I need to replace the words that have parentheses. I want to replace "(happy)" with "good", and "(explore)" with "learn".
I have some ideas, but I don't know how.
for (int i = 0; i <= string.length(), i++) {
for (int j = 0; j <= string.length(), j++
if ((string.charAt(i)== '(') && (string.charAt(j) == ')')) {
String w1 = line.substring(i+1,j);
string.replace(w1, w2)
}
}
}
My problem is that I can only replace one word with one new word...
I am thinking of using a scanner to prompt me to give a new word and then replace it, how can I do this?
The appendReplacement and appendTail methods of Matcher are designed for this purpose. You can use a regex to scan for your pattern--a pair of parentheses with a word in the middle--then do whatever you need to do to determine the string to replace it with. See the javadoc.
An example, based on the example in the javadoc. I'm assuming you have two methods, replacement(word) that tells what you want to replace the word with (so that replacement("happy") will equal "good" in your example), and hasReplacement(word) that tells whether the word has a replacement or not.
Pattern p = Pattern.compile("\\((.*?)\\)");
Matcher m = p.matcher(source);
StringBuffer sb = new StringBuffer();
while (m.find()) {
String word = m.group(1);
String newWord = hasReplacement(word) ? replacement(word) : m.group(0);
m.appendReplacement(sb, newWord); // appends the replacement, plus any not-yet-used text that comes before the match
}
m.appendTail(sb); // appends any text left over after the last match
String result = sb.toString();
Use below code for replacing the string.
String string = "today is a (happy) day, I would like to (explore) more about Java.";
string = string.replaceAll("\\(happy\\)", "good");
string = string.replaceAll("\\(explore\\)", "learn");
System.out.println(string);`
What you can do is run a loop from 0 to length-1 and if loop encounters a ( then assign its index to a temp1 variable. Now go on further as long as you encounter ).Assign its index to temp2 .Now you can replace that substring using string.replace(string.substring(temp1+1,temp2),"Your desired string")).
No need to use the nested loops. Better use one loop and store the index when you find opening parenthesis and also for close parenthesis and replace it with the word. Continue the same loop and store next index. As you are replacing the words in same string it changes the length of string you need to maintain copy of string and perform loop and replace on different,
Do not use nested for loop. Search for occurrences of ( and ). Get the substring between these two characters and then replace it with the user entered value. Do it till there are not more ( and ) combinations left.
import java.util.Scanner;
public class ReplaceWords {
public static String replaceWords(String s){
while(s.contains(""+"(") && s.contains(""+")")){
Scanner keyboard = new Scanner(System.in);
String toBeReplaced = s.substring(s.indexOf("("), s.indexOf(")")+1);
System.out.println("Enter the word with which you want to replace "+toBeReplaced+" : ");
String replaceWith = keyboard.nextLine();
s = s.replace(toBeReplaced, replaceWith);
}
return s;
}
public static void main(String[] args) {
String myString ="today is a (happy) day, I would like to (explore) more about Java.";
myString = replaceWords(myString);
System.out.println(myString);
}
}
This snippet works for me, just load the HashMap up with replacements and iterate through:
import java.util.*;
public class Test
{
public static void main(String[] args) {
String string = "today is a (happy) day, I would like to (explore) more about Java.";
HashMap<String, String> hm = new HashMap<String, String>();
hm.put("\\(happy\\)", "good");
hm.put("\\(explore\\)", "learn");
for (Map.Entry<String, String> entry : hm.entrySet()) {
String key = entry.getKey();
String value = entry.getValue();
string = string.replaceAll(key, value);
}
System.out.println(string);
}
}
Remember, replaceAll takes a regex, so you want it to display "\(word\)", which means the slashes themselves must be escaped.

How can I eliminate duplicate words from String in Java?

I have an ArrayList of Strings and it contains records such as:
this is a first sentence
hello my name is Chris
what's up man what's up man
today is tuesday
I need to clear this list, so that the output does not contain repeated content. In the case above, the output should be:
this is a first sentence
hello my name is Chris
what's up man
today is tuesday
as you can see, the 3rd String has been modified and now contains only one statement what's up man instead of two of them.
In my list there is a situation that sometimes the String is correct, and sometimes it is doubled as shown above.
I want to get rid of it, so I thought about iterating through this list:
for (String s: myList) {
but I cannot find a way of eliminating duplicates, especially since the length of each string is not determined, and by that I mean there might be record:
this is a very long sentence this is a very long sentence
or sometimes short ones:
single word singe word
is there some native java function for that maybe?
Assuming the String is repeated just twice, and with an space in between as in your examples, the following code would remove repetitions:
for (int i=0; i<myList.size(); i++) {
String s = myList.get(i);
String fs = s.substring(0, s.length()/2);
String ls = s.substring(s.length()/2+1, s.length());
if (fs.equals(ls)) {
myList.set(i, fs);
}
}
The code just split each entry of the list into two substrings (dividing by the half point). If both are equal, substitute the original element with only one half, thus removing the repetition.
I was testing the code and did not see #Brendan Robert answer. This code follows the same logic as his answer.
I would suggest using regular expressions. I was able to remove duplicates using this pattern: \b([\w\s']+) \1\b
public class Main {
static String [] phrases = {
"this is a first sentence",
"hello my name is Chris",
"what's up man what's up man",
"today is tuesday",
"this is a very long sentence this is a very long sentence",
"single word single word",
"hey hey"
};
public static void main(String[] args) throws Exception {
String duplicatePattern = "\\b([\\w\\s']+) \\1\\b";
Pattern p = Pattern.compile(duplicatePattern);
for (String phrase : phrases) {
Matcher m = p.matcher(phrase);
if (m.matches()) {
System.out.println(m.group(1));
} else {
System.out.println(phrase);
}
}
}
}
Results:
this is a first sentence
hello my name is Chris
what's up man
today is tuesday
this is a very long sentence
single word
hey
Assumptions:
Uppercase words are equal to lowercase counterparts.
String fullString = "lol lol";
String[] words = fullString.split("\\W+");
StringBuilder stringBuilder = new StringBuilder();
Set<String> wordsHashSet = new HashSet<>();
for (String word : words) {
// Check for duplicates
if (wordsHashSet.contains(word.toLowerCase())) continue;
wordsHashSet.add(word.toLowerCase());
stringBuilder.append(word).append(" ");
}
String nonDuplicateString = stringBuilder.toString().trim();
simple logic : split every word by token space i.e " " and now add it in LinkedHashSet , Retrieve back, Replace "[","]",","
String s = "I want to walk my dog I want to walk my dog";
Set<String> temp = new LinkedHashSet<>();
String[] arr = s.split(" ");
for ( String ss : arr)
temp.add(ss);
String newl = temp.toString()
.replace("[","")
.replace("]","")
.replace(",","");
System.out.println(newl);
o/p : I want to walk my dog
It depends on the situation that you have but assuming that the string can be repeated at most twice and not three or more times you could find the length of the entire string, find the halfway point and compare each index after the halfway point with the matching beginning index. If the string can be repeated more than once you will need a more complicated algorithm that would first determine how many times the string is repeated and then finds the starting index of each repeat and truncates all index's from the beginning of the first repeat onward. If you can provide some more context into what possible scenarios you expect to handle we can start putting together some ideas.
//Doing it in Java 8
String str1 = "I am am am a good Good coder";
String[] arrStr = str1.split(" ");
String[] element = new String[1];
return Arrays.stream(arrStr).filter(str1 -> {
if (!str1.equalsIgnoreCase(element[0])) {
element[0] = str1;
return true;
}return false;
}).collect(Collectors.joining(" "));

Convert String into Title Case

I am a beginner in Java trying to write a program to convert strings into title case. For example, if String s = "my name is milind", then the output should be "My Name Is Milind".
import java.util.*;
class TitleCase
{
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
System.out.println("ent");
String s=in.nextLine();
String str ="";
char a ;
for(int i =0;i<s.length()-1;i++)
{
a = s.charAt(i);
if(a==' ')
{
str = str+(Character.toUpperCase(s.charAt(i+1)));
}
else
{
str =str+(Character.toLowerCase(a));
}
}
//for(int i =0; i<s.length();i++)
//{
System.out.println(str);
//}
}
}
You are trying to capitalize every word of the input.
So you have to do following steps:
get the words separated
capitalize each word
put it all together
print it out
Example Code:
public static void main(String args[]){
Scanner in = new Scanner(System.in);
System.out.println("ent");
String s=in.nextLine();
//now your input string is storred inside s.
//next we have to separate the words.
//here i am using the split method (split on each space);
String[] words = s.split(" ");
//next step is to do the capitalizing for each word
//so use a loop to itarate through the array
for(int i = 0; i< words.length; i++){
//we will save the capitalized word in the same place again
//first, geht the character on first position
//(words[i].charAt(0))
//next, convert it to upercase (Character.toUppercase())
//then add the rest of the word (words[i].substring(1))
//and store the output back in the array (words[i] = ...)
words[i] = Character.toUpperCase(words[i].charAt(0)) +
[i].substring(1);
}
//now we have to make a string out of the array, for that we have to
// seprate the words with a space again
//you can do this in the same loop, when you are capitalizing the
// words!
String out = "";
for(int i = 0; i<words.length; i++){
//append each word to out
//and append a space after each word
out += words[i] + " ";
}
//print the result
System.out.println(out);
}
Using Java 8 streams:
String titleCase = (new ArrayList<>(Arrays.asList(inputString.toLowerCase().split(" "))))
.stream()
.map(word -> Character.toTitleCase(word.charAt(0)) + word.substring(1))
.collect(Collectors.joining(" "));
The problem is with the way you're adding characters. Take a look at your if condition:
a = s.charAt(i);
if(a==' ')
{
// Here you are adding not the current character, but the NEXT character.
str = str+(Character.toUpperCase(s.charAt(i+1)));
}
else
{
// Here you are adding the current character.
str =str+(Character.toLowerCase(a));
}
As a result of this condition, you will skip a character if your input string contains a space, then repeat another character that you've already added.
Additionally, you're not looping through the whole string because your loop conditional goes to s.length()-1. Change that to just s.length(). However, if you do that, you may run into an exception if the input string ends with a space (since you'll try to check for a character at an out-of-bound index).
Here's what the fixed code would look like:
public static void main(String args[])
{
Scanner in = new Scanner(System.in);
System.out.println("ent");
String s=in.nextLine();
String str ="";
char a ;
for(int i =0;i<s.length();i++)
{
a = s.charAt(i);
if(a==' ')
{
str = str+Character.toLowerCase(a)+(Character.toUpperCase(s.charAt(i+1)));
i++; // "skip" the next element since it is now already processed
}
else
{
str =str+(Character.toLowerCase(a));
}
}
System.out.println(str);
}
NOTE: I only fixed the code that you supplied. However, I'm not sure it works the way you want it to - the first character of the string will still be whatever case it started in. Your conditional only uppercases letters that are preceded by a space.
You want to change the case of the first letter of each word of a String.
To do so, I would follow the following steps :
split the String in words : see String.split(separator)
retrieve the first letter of each word : see String.charAt(index)
retrieve its capitalized version : the Character.toUpperCase(char) you use is perfect
concatenate the capitalized letter with the rest of the word : concatenation operator (+) and String.substring
create a new String from the capitalized words : see String.join(separator)
Code Golf variation... I challenge anyone to make it any simpler than this:
public String titleCase(String str) {
return Arrays
.stream(str.split(" "))
.map(String::toLowerCase)
.map(StringUtils::capitalize)
.collect(Collectors.joining(" "));
}
By the way: Unicode distinguishes between three cases: lower case, upper case and title case. Although it does not matter for English, there are other languages where the title case of a character does not match the upper case version. So you should use
Character.toTitleCase(ch)
instead of Character.toUpperCase(ch) for the first letter.
There are three character cases in Unicode: upper, lower, and title. Uppercase and lowercase are familiar to most people. Titlecase distinguishes characters that are made up of multiple components and are written differently when used in titles, where the first letter in a word is traditionally capitalized. For example, in the string "ljepotica",[2] the first letter is the lowercase letter lj(\u01C9 , a letter in the Extended Latin character set that is used in writing Croatian digraphs). If the word appeared in a book title, and you wanted the first letter of each word to be in uppercase, the correct process would be to use toTitleCase on the first letter of each word, giving you "Ljepotica" (using Lj, which is \u01C8). If you incorrectly used toUpperCase, you would get the erroneous string "LJepotica" (using LJ, which is \u01C7).
[The Java™ Programming Language, Fourth Edition, by James Gosling, Ken Arnold, David Holmes (Prentice Hall). Copyright 2006 Sun Microsystems, Inc., 9780321349804]
WordUtils.capitalizeFully() worked for me like charm as it gives: WordUtils.capitalizeFully("i am FINE") = "I Am Fine"
import java.util.Scanner;
public class TitleCase {
public static void main(String[] args) {
System.out.println("please enter the string");
Scanner sc1 = new Scanner(System.in);
String str = sc1.nextLine();
//whatever the format entered by user, converting it into lowercase
str = str.toLowerCase();
// converting string to char array for
//performing operation on individual elements
char ch[] = str.toCharArray();
System.out.println("===============");
System.out.println(str);
System.out.println("===============");
//First letter of senetence must be uppercase
System.out.print((char) (ch[0] - 32));
for (int i = 1; i < ch.length; i++) {
if (ch[i] == ' ') {
System.out.print(" " + (char) (ch[i + 1] - 32));
//considering next variable after space
i++;
continue;
}
System.out.print(ch[i]);
}
}
}
You can use lamda instead-
String titalName = Arrays.stream(names.split(" "))
.map(E -> String.valueOf(E.charAt(0))+E.substring(1))
.reduce(" ", String::concat);

Java String.contains() to take care of natural numbers

I'm a computer science student learning Java, and as an exercise, we're doing a permutation algorhythm.
Now, i'm stuck at a point where i need to search for a natural number within a String full of numbers, splitted by a comma:
String myString = "0,1,2,10,14,";
The problem is i'm using...
myString.contains(String.valueOf(anInteger);
...to check for the presence of a specific number. This works for numbers from 0 to 9, but when looking for a more-than-1-digit number, the program does not recognize it as a natural number.
In other words, and as an example: "14" is not the integer 14, its just a string with an "1", and a "4"; so, if i run...
String myString = "0,1,2,10,14,";
if (myString.contains(myString.valueOf(4))) { doSomething(); }
...the "if" statement will be true, since the integer "4" is present in the string, as part of the natural number "14".
At this point, i've been searching through StackOverflow and other pages for a solution, and learnt i should use Pattern and Matcher.
My question is: what's the best way to do use them?
Relevant part of my code:
for (int i = 0; i<r; i++)
{
if (!act.contains(String.valueOf(i)))
{
...
}
...
}
I use this method several times in my code, so an exact substitution would be nice.
Thank you all in advance!
You only need a method call to matches():
if (myString.matches(".*\\b" + anInteger + "\\b.*"))
// string contains the number
This works using by creating a regex that has a word boundary (\b) at either end of the target number. The leading and trailing .* are required because matches() must match the whole string to return true.
Look into how to split a String into an array of String. So:
String[] splitStrings = myString.split(",")
ArrayList<Integer> parsedInts = new ArrayList<Integer>();
for (String str : splitStrings) {
parsedInts.add(Integer.parseInt(str));
}
then in your for loop:
if (parsedInts.contains(i)) {
// body
}
Something like this:
String myString = "0,1,2,10,14,";
String[] split = myString.split(",");
for (String string : split) {
int num = Integer.parseInt(string);
if (num == 4) {
System.out.println(num);
// ...
}
}
String myString = "0,1,2,10,14,2323232";
String[] allList = myString.split(",");
for (String string : allList) {
if(string.matches("[0-9]*"))
{
System.out.println("Its number with value "+string);
}
}
I think you need to pick all the numbers in the given string and find the permutation.
I think you need to Tokenize the given string with the Comma Separator.
When I do such program, I divide my logic to parse the String and write the logic in another method. Below is the snippet
String myString = "0,1,2,10,14,";
StringTokenizer st2 = new StringTokenizer(myString , ",");
while (st2.hasMoreElements()) {
doSomething(st2.nextElement());
}

How to check if all characters in a String are all letters?

I'm able to separate the words in the sentence but I do not know how to check if a word contains a character other than a letter. You don't have to post an answer just some material I could read to help me.
public static void main(String args [])
{
String sentance;
String word;
int index = 1;
System.out.println("Enter sentance please");
sentance = EasyIn.getString();
String[] words = sentance.split(" ");
for ( String ss : words )
{
System.out.println("Word " + index + " is " + ss);
index++;
}
}
What I would do is use String#matches and use the regex [a-zA-Z]+.
String hello = "Hello!";
String hello1 = "Hello";
System.out.println(hello.matches("[a-zA-Z]+")); // false
System.out.println(hello1.matches("[a-zA-Z]+")); // true
Another solution is if (Character.isLetter(str.charAt(i)) inside a loop.
Another solution is something like this
String set = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ";
String word = "Hello!";
boolean notLetterFound;
for (char c : word.toCharArray()){ // loop through string as character array
if (!set.contains(c)){ // if a character is not found in the set
notLetterfound = true; // make notLetterFound true and break the loop
break;
}
}
if (notLetterFound){ // notLetterFound is true, do something
// do something
}
I prefer the first answer though, using String#matches
For more reference goto-> How to determine if a String has non-alphanumeric characters?
Make the following changes in pattern "[^a-zA-Z^]"
Not sure if I understand your question, but there is the
Character.isAlpha(c);
You would iterate over all characters in your string and check whether they are alphabetic (there are other "isXxxxx" methods in the Character class).
You could loop through the characters in the word calling Character.isLetter(), or maybe check if it matches a regular expression e.g. [\w]* (this would match the word only if its contents are all characters).
you can use charector array to do this like..
char[] a=ss.toCharArray();
not you can get the charector at the perticulor index.
with "word "+index+" is "+a[index];

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