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Extraction of numerals from a string

I need to extract the numerals in a given string and store them in a separate array such that each index stores a individual number in the string.
Ex-"15 foxes chases 12 rabbits". I need the numbers 15 and 12 to be stored in a[0] and a[1].
String question=jTextArea1.getText();
String lower=question.toLowerCase();
check(lower);
public void check(String low)
{
int j;
String[] ins={"into","add","Insert"};
String cc=low;
for(int i=0;i<2;i++)
{
String dd=ins[i];
if(cc.contains(dd))
{
j=1;
insert(cc);
break;
}
}}
public void insert(String low)
{
String character = low;
int l=low.length();
int j[]=new int[20];
int m=0;
for(int k=0;k<=2;k++)
{
j[k]=0;
for(int i=0;i<l;i++)
{
char c = character.charAt(i);
if (Character.isDigit(c))
{
String str=Character.toString(c);
j[k]=(j[k]*10)+Integer.parseInt(str);
m++;
}
else if (Character.isLetter(c))
{
if(m>2)
{
break;
}
}}}

Regex is the best option for you.
//Compilation of the regex
Pattern p = Pattern.compile("(\d*)");
// Creation of the search engine
Matcher m = p.matcher("15 foxes chases 12 rabbits");
// Lunching the searching
boolean b = m.matches();
// if any number was found then
if(b) {
// for each found number
for(int i=1; i <= m.groupCount(); i++) {
// Print out the found numbers;
// if you want you can store these number in another array
//since m.group is the one which has the found number(s)
System.out.println("Number " + i + " : " + m.group(i));
}
}
You must import java.util.regex.*;

Assuming you can't use a Collection like a List, I would start by splitting on white space, using a method to count the numbers in that array (with a Pattern on digits), and then build the output array. Something like
public static int getNumberCount(String[] arr) {
int count = 0;
for (String str : arr) {
if (str.matches("\\d+")) {
count++;
}
}
return count;
}
And then use it like
public static void main(String[] args) {
String example = "15 foxes chases 12 rabbits";
String[] arr = example.split("\\s+");
int count = getNumberCount(arr);
int[] a = new int[count];
int pos = 0;
for (String str : arr) {
if (str.matches("\\d+")) {
a[pos++] = Integer.parseInt(str);
}
}
System.out.println(Arrays.toString(a));
}
Output is (as requested)
[15, 12]

To extract numerals from Strings use the following code:
String text = "12 18 100";
String[] n = text.split(" ");
int[] num = n.length;
for (int i =0; i < num.length;i++) {
num[i] = Integer.parseInt(n[i]);
};
All the string numbers in the text will be then integers

I believie it should be
String str = "12 aa 15 155";
Scanner scanner = new Scanner( str );
while( scanner.hasNext() )
{
if( scanner.hasNextInt() )
{
int next = scanner.nextInt();
System.out.println( next );
}
else
{
scanner.next();
}
}
scanner.close();

Java words reverse

I am new to Java and I found a interesting problem which I wanted to solve. I am trying to code a program that reverses the position of each word of a string. For example, the input string = "HERE AM I", the output string will be "I AM HERE". I have got into it, but it's not working out for me. Could anyone kindly point out the error, and how to fix it, because I am really curious to know what's going wrong. Thanks!
import java.util.Scanner;
public class Count{
static Scanner sc = new Scanner(System.in);
static String in = ""; static String ar[];
void accept(){
System.out.println("Enter the string: ");
in = sc.nextLine();
}
void intArray(int words){
ar = new String[words];
}
static int Words(String in){
in = in.trim(); //Rm space
int wc = 1;
char c;
for (int i = 0; i<in.length()-1;i++){
if (in.charAt(i)==' '&&in.charAt(i+1)!=' ') wc++;
}
return wc;
}
void generate(){
char c; String w = ""; int n = 0;
for (int i = 0; i<in.length(); i++){
c = in.charAt(i);
if (c!=' '){
w += c;
}
else {
ar[n] = w; n++;
}
}
}
void printOut(){
String finale = "";
for (int i = ar.length-1; i>=0;i--){
finale = finale + (ar[i]);
}
System.out.println("Reversed words: " + finale);
}
public static void main(String[] args){
Count a = new Count();
a.accept();
int words = Words(in);
a.intArray(words);
a.generate();
a.printOut();
}
}

Got it. Here is my code that implements split and reverse from scratch.
The split function is implemented through iterating through the string, and keeping track of start and end indexes. Once one of the indexes in the string is equivalent to a " ", the program sets the end index to the element behind the space, and adds the previous substring to an ArrayList, then creating a new start index to begin with.
Reverse is very straightforward - you simply iterate from the end of the string to the first element of the string.
Example:
Input: df gf sd
Output: sd gf df
import java.util.Scanner;
import java.util.ArrayList;
import java.util.Collections;
public class Count{
public static void main(String[] args)
{
Scanner scan = new Scanner(System.in);
System.out.println("Enter string to reverse: ");
String unreversed = scan.nextLine();
System.out.println("Reversed String: " + reverse(unreversed));
}
public static String reverse(String unreversed)
{
ArrayList<String> parts = new ArrayList<String>();
String reversed = "";
int start = 0;
int end = 0;
for (int i = 0; i < unreversed.length(); i++)
{
if (unreversed.charAt(i) == ' ')
{
end = i;
parts.add(unreversed.substring(start, end));
start = i + 1;
}
}
parts.add(unreversed.substring(start, unreversed.length()));
for (int i = parts.size()-1; i >= 0; i--)
{
reversed += parts.get(i);
reversed += " ";
}
return reversed;
}
}

There is my suggestion :
String s = " HERE AM I ";
s = s.trim();
int j = s.length() - 1;
int index = 0;
StringBuilder builder = new StringBuilder();
for (int i = j; i >= 0; i--) {
Character c = s.charAt(i);
if (c.isWhitespace(c)) {
index = i;
String r = s.substring(index+1, j+1);
j = index - 1;
builder.append(r);
builder.append(" ");
}
}
String r=s.substring(0, index);
builder.append(r);
System.out.println(builder.toString());

From adding debug output between each method call it's easy to determine that you're successfully reading the input, counting the words, and initializing the array. That means that the problem is in generate().
Problem 1 in generate() (why "HERE" is duplicated in the output): after you add w to your array (when the word is complete) you don't reset w to "", meaning every word has the previous word(s) prepended to it. This is easily seen by adding debug output (or using a debugger) to print the state of ar and w each iteration of the loop.
Problem 2 in generate() (why "I" isn't in the output): there isn't a trailing space in the string, so the condition that adds a word to the array is never met for the last word before the loop terminates at the end of the string. The easy fix is to just add ar[n] = w; after the end of the loop to cover the last word.

I would use the split function and then print from the end of the list to the front.
String[] splitString = str.split(" ");
for(int i = splitString.length() - 1; i >= 0; i--){
System.out.print(splitString[i]);
if(i != 0) System.out.print(' ');
}
Oops read your comment. Disregard this if it is not what you want.

This has a function that does the same as split, but not the predefined split function
public static void main(String args[]) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter the string : ");
String input = sc.nextLine();
// This splits the string into array of words separated with " "
String arr[] = myOwnSplit(input.trim(), ' '); // ["I", "AM", "HERE"]
// This ll contain the reverse string
String rev = "";
// Reading the array from the back
for(int i = (arr.length - 1) ; i >= 0 ; i --) {
// putting the words into the reverse string with a space to it's end
rev += (arr[i] + " ");
}
// Getting rid of the last extra space
rev.trim();
System.out.println("The reverse of the given string is : " + rev);
}
// The is my own version of the split function
public static String[] myOwnSplit(String str, char regex) {
char[] arr = str.toCharArray();
ArrayList<String> spltedArrayList = new ArrayList<String>();
String word = "";
// splitting the string based on the regex and bulding an arraylist
for(int i = 0 ; i < arr.length ; i ++) {
char c = arr[i];
if(c == regex) {
spltedArrayList.add(word);
word = "";
} else {
word += c;
}
if(i == (arr.length - 1)) {
spltedArrayList.add(word);
}
}
String[] splitedArray = new String[spltedArrayList.size()];
// Converting the arraylist to string array
for(int i = 0 ; i < spltedArrayList.size() ; i++) {
splitedArray[i] = spltedArrayList.get(i);
}
return splitedArray;
}

Split a String into number of Characters desired by the user

I want to split a String into n number of characters.
Consider input to be "Example-for-my-Question". Now if I want to split into n=3 characters, output should be "Exa, mpl, e-f, or-, my-, Que, sti, on" and suppose n=4, output should be "Exam, ple-, for-, my-Q, uest, ion" How can you modify the program below without using POSIX.
import java.util.Scanner;
public class SplitString {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
System.out.println("Enter a String; ");
String inputString = in.nextLine();
System.out.println("How many characters do you want to split into ?");
int n = in.nextInt();
String[] array = inputString.split(" ", n);
System.out.println("Number of words: " + array.length);
for (String arr : array)
System.out.println(arr);
}
}

The simple way to do this is to use String.substring(...) repeatedly to trim N characters off the front of your string ... in a loop.
But if you really want to do this using String.split(...), then I think that the separator regex needs to be a positive look-behind that matches N characters. (It is obscure, and inefficient ... but if regexes are your universal tool ...)

You can use substring for this task.
String sp="StackOverFlow";
int NoOfChars=3;
for(int i=0;i<sp.length();i+=NoOfChars)
{
if(i+NoOfChars<=sp.length())
System.out.println(sp.substring(i,i+NoOfChars));
//Instead add in String ArrayList
else
System.out.println(sp.substring(i));
}
OUTPUT
Sta
ckO
ver
Flo
w
NOTE:Better to use trim() to remove leading or trailing spces

This works for me. In addition to splitting into known lengths, it checks for a null or "too small of a" source string, etc. If a null string is supplied, then a null is returned. If the source string is smaller than the requested split length, then the source string is simply returned.
public static void main (String[] args) throws java.lang.Exception
{
// Three test cases...
String pieces[] = SplitString("Example-for-my-Question", 3);
//String pieces[] = SplitString("Ex", 3);
//String pieces[] = SplitString(null, 3);
if (null != pieces)
{
for (int i = 0; i < pieces.length; i++)
{
System.out.println(pieces[i]);
}
}
}
private static String[] SplitString(String source, int size)
{
String result[] = null;
if (null != source && source.length() > size)
{
int numberOfElements = source.length() / size;
int modulo = source.length() % size;
if (modulo > 0)
{
numberOfElements++;
}
result = new String[numberOfElements];
for (int i = 0; i < numberOfElements; i++)
{
if (numberOfElements - 1 != i)
{
result[i] = source.substring(i * size, (i * size) + size);
}
else
{
result[numberOfElements - 1] = source.substring(i * size, (i * size) + modulo);
}
}
}
else if (null != source)
{
result = new String[1];
result[0] = source;
}
return result;
}

Please try the following program, but here you have to give input to "N" inside the program itself
class Main {
public static void main(String[] args) {
int N = 5;
String text = "aaaaabbbbbccccceeeeefff";
String[] tokens = text.split("(?<=\\G.{" + N + "})");
for(String t : tokens) {
System.out.println(t);
}
}
}

Generating the lexicographically greatest string

The question is to generate the lexicographically greatest string given some string s.
So the aim is to find lexicographically greatest, unique(no repetitions) substring s1 from s.
We say that some subsequence s1 is greater than another subsequence s2 if s1 has more characters than s2 or s1 is lexicographically greater than s2 if equal length.
I/O are as follows:
Input is: babab
output is: ba
Second input is: nlhthgrfdnnlprjtecpdrthigjoqdejsfkasoctjijaoebqlrgaiakfsbljmpibkidjsrtkgrdnqsknbarpabgokbsrfhmeklrle
Second output is:
tsocrpkijgdqnbafhmle
This is what I wrote for my java code but my code fails on the second test case. Also I'm having a hard time understanding why second output isn't tsrqponmlkjihgfedcba.
Can somebody provide suggestions for a fix or even java code?
I think the algorithm has to be more efficient than generating all possible unique strings, sort them and find lexicographically largest one.
To make the question much clearer, if the input is babab, then all the possible unique combinations would be b, a, ba, ab. And the output will be ba because it's the longest and lexicographically greater than ab.
Note: this is not a homework assignment.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class mostBeautiful {
final static int MAX = 1000000;
static String[] permute;
static void permutation(String prefix, String str, int counter) {
int n = str.length();
//System.out.println("n is: "+ n);
if (n == 0) {
permute[counter] = prefix;
} else {
for (int i = 0; i < n; i++) {
//System.out.println("str is: "+ str);
permutation(prefix + str.charAt(i), str.substring(0, i) + str.substring(i+1, n), counter++);
}
}
}
public static void main(String[] args) throws IOException {
BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
String s = bf.readLine();
char[] unique = new char[26];
int counter = 0;
String answer = "";
//System.out.println("s is: " + s);
int ascii = 0;
final int asciiAVal = 97;
final int asciiZVal = 122;
for (int i = 0; i < s.length(); i++) {
ascii = (int)s.charAt(i);
if (ascii < asciiAVal || ascii > asciiZVal) {
continue;
}
char ch = s.charAt(i);
unique[ch - 'a'] = ch;
}
String result = "";
for (int j = 25; j >= 0; j--) {
result += unique[j];
}
result = result.trim();
System.out.println(result);
int size = result.length() * (result.length() - 1);
permute = new String[size];
permutation("", result, counter);
for (int i = 1; i < size; i++) {
if (permute[i].compareTo(permute[i - 1]) > 0){
answer = permute[i];
} else {
answer = permute[i - 1];
}
}
System.out.println("answer is: " + answer);
}
}

After thinking about this problem in many ways, I have determined a divide-and-conquer algorithm that gets the results right:
Algorithm - Pseudocode
Assuming some input string, S defined as a concatenation of two substrings A + B, we compute the lexicographically greatest string recursively as:
LexMax(S) = Merge(LexMax(A),LexMax(B))
Where
LexMax(S)
{
if Length(S) = 1
return S
else
{
LMA = LexMax(S[0:Length/2])
LMB = LexMax(S[Length/2:end])
return Merge(LMA,LMB)
}
}
Merge(A,B)
{
Sa = A
Sb = B
for n = 0:Length(A)
{
if Sb contains A[n]
{
if A[n+1:end] contains character > A[n]
Remove A[n] from Sa
else
Remove A[n] from Sb
}
}
return Sa + Sb
}
Java Code
Coming soon!
Example
Given an input string
cefcfdabbcfed
Divide it into
cefcfda
bbcfed
Assuming the function works we have:
LexMax("cefcfda") = "efcda"
LexMax("bbcfed") = "bcfed"
Merging works as follows:
e: efcda bcfed
In both substrings, greater value found to right of e in left substring, remove from left
f: fcda bcfed
In both substrings, no greater value in left substring, remove from right
c: fcda bced
In both substrings, greater value found to right of c in left substring, remove from left
d: fda bced
In both substrings, no greater value in left substring, remove from right
a: fda bce
Not in both substrings, do nothing
Final result:
LexMax(cefcfdabbcfed) = fdabce

This is not a direct answer, but doesn't this code meet the requirement as you explained it in the discussion above?
final String x = "saontehusanoethusnaoteusnaoetuh";
final SortedSet<Character> chars =
new TreeSet<Character>(Collections.reverseOrder());
for (char c : x.toCharArray()) chars.add(c);
System.out.println(chars);

Lexicographic order is an order in which words are displayed in alphabetical order using the appearance of letters in the word.It is also know as dictionary order or alphabetical order.For ex:-"Africa" is smaller than "Bangladesh" ,"He" is smaller than "he".
public class LexicographicExample {
public static void main(String a[]) {
Scanner sc = new Scanner(System.in);
System.out.println("Enter the String:-");
String str = sc.nextLine();
System.out.println("Enter the length");
int count = sc.nextInt();
List<String> list = new ArrayList<String>();
for (int i = 0; i < str.length(); i = i + 1) {
if (str.length() - i >= count) {
list.add(str.substring(i, count + i));
}
}
Collections.sort(list);
System.out.println("Smallest subString:-" + list.get(0));
System.out.println("Largest subString:-" + list.get(list.size() - 1));
}
}
For reference ,refer this link http://techno-terminal.blogspot.in/2015/09/java-program-to-find-lexicographically.html

"tsrqponmlkjihgfedcba" is not the answer because it is not a subsequence of the input. The definition of subsequence requires that the characters of the subsequence occur in the original sequence in the same order. For example, "abc" is a subsequence of "apbqcr", while "cba" is not.
As to the solution, I think a simple greedy algorithm would suffice. First, one has to understand that the maximum possible length of the output is the number of unique symbols (say, N) in the input. Since any output shorter than that would not be the greatest one, it has to be exactly N symbols long. The rest of the procedure is simple and at most quadratic in time complexity: one has to go through the input string and at each step pick the lexicographically highest symbol such that the part of the string to the left of it would still contain all the "unused" symbols.
As an example, consider a string "bacb". The first symbol can be 'a' or 'b', since in both cases the remainder contains both of the other letters. 'b' is greater, so we pick it. Now for "acb" we can only pick 'a' and than 'c' according to that condition, so we end up with "bac" for output.

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Scanner;
class aaa {
public static void main(String args[]) throws Exception {
Scanner scan = new Scanner(System.in);
// int n = scan.nextInt();
String s = scan.next();
HashMap<Character, Node5> map = new HashMap<>();
for (int i = 0; i < s.length(); i++) {
if (!map.containsKey(s.charAt(i))) {
Node5 node = new Node5();
node.nl.add(i);
node.li = i;
map.put(s.charAt(i), node);
} else {
Node5 rn = map.get(s.charAt(i));
rn.nl.add(i);
rn.li = i;
map.put(s.charAt(i), rn);
}
}
String s1 = "";
int index = -1;
for (int i = 25; i >= 0; i--) {
if (map.containsKey((char) (97 + i))) {
if (map.get((char) (97 + i)).li > index) {
for (int j = 0; j < map.get((char) (97 + i)).nl.size(); j++) {
if (map.get((char) (97 + i)).nl.get(j) > index) {
s1 += (char) (97 + i);
index = map.get((char) (97 + i)).nl.get(j);
}
}
}
}
}
System.out.println(s1);
scan.close();
}
}
class Node5 {
int li;
ArrayList<Integer> nl;
public Node5() {
this.nl = new ArrayList<>();
}
}

Java: method to get position of a match in a String?

String match = "hello";
String text = "0123456789hello0123456789";
int position = getPosition(match, text); // should be 10, is there such a method?

The family of methods that does this are:
int indexOf(String str)
indexOf(String str, int fromIndex)
int lastIndexOf(String str)
lastIndexOf(String str, int fromIndex)
Returns the index within this string of the first (or last) occurrence of the specified substring [searching forward (or backward) starting at the specified index].
String text = "0123hello9012hello8901hello7890";
String word = "hello";
System.out.println(text.indexOf(word)); // prints "4"
System.out.println(text.lastIndexOf(word)); // prints "22"
// find all occurrences forward
for (int i = -1; (i = text.indexOf(word, i + 1)) != -1; i++) {
System.out.println(i);
} // prints "4", "13", "22"
// find all occurrences backward
for (int i = text.length(); (i = text.lastIndexOf(word, i - 1)) != -1; i++) {
System.out.println(i);
} // prints "22", "13", "4"

This works using regex.
String text = "I love you so much";
String wordToFind = "love";
Pattern word = Pattern.compile(wordToFind);
Matcher match = word.matcher(text);
while (match.find()) {
System.out.println("Found love at index "+ match.start() +" - "+ (match.end()-1));
}
Output :
Found 'love' at index 2 - 5
General Rule :
Regex search left to right, and once the match characters has been used, it cannot be reused.

text.indexOf(match);
See the String javadoc

Finding a single index
As others have said, use text.indexOf(match) to find a single match.
String text = "0123456789hello0123456789";
String match = "hello";
int position = text.indexOf(match); // position = 10
Finding multiple indexes
Because of #StephenC's comment about code maintainability and my own difficulty in understanding #polygenelubricants' answer, I wanted to find another way to get all the indexes of a match in a text string. The following code (which is modified from this answer) does so:
String text = "0123hello9012hello8901hello7890";
String match = "hello";
int index = text.indexOf(match);
int matchLength = match.length();
while (index >= 0) { // indexOf returns -1 if no match found
System.out.println(index);
index = text.indexOf(match, index + matchLength);
}

You can get all matches in a file simply by assigning inside while-loop, cool:
$ javac MatchTest.java
$ java MatchTest
1
16
31
46
$ cat MatchTest.java
import java.util.*;
import java.io.*;
public class MatchTest {
public static void main(String[] args){
String match = "hello";
String text = "hello0123456789hello0123456789hello1234567890hello3423243423232";
int i =0;
while((i=(text.indexOf(match,i)+1))>0)
System.out.println(i);
}
}

int match_position=text.indexOf(match);

import java.util.StringTokenizer;
public class Occourence {
public static void main(String[] args) {
String key=null,str ="my name noorus my name noorus";
int i=0,tot=0;
StringTokenizer st=new StringTokenizer(str," ");
while(st.hasMoreTokens())
{
tot=tot+1;
key = st.nextToken();
while((i=(str.indexOf(key,i)+1))>0)
{
System.out.println("position of "+key+" "+"is "+(i-1));
}
}
System.out.println("total words present in string "+tot);
}
}

I have some big code but working nicely....
class strDemo
{
public static void main(String args[])
{
String s1=new String("The Ghost of The Arabean Sea");
String s2=new String ("The");
String s6=new String ("ehT");
StringBuffer s3;
StringBuffer s4=new StringBuffer(s1);
StringBuffer s5=new StringBuffer(s2);
char c1[]=new char[30];
char c2[]=new char[5];
char c3[]=new char[5];
s1.getChars(0,28,c1,0);
s2.getChars(0,3,c2,0);
s6.getChars(0,3,c3,0); s3=s4.reverse();
int pf=0,pl=0;
char c5[]=new char[30];
s3.getChars(0,28,c5,0);
for(int i=0;i<(s1.length()-s2.length());i++)
{
int j=0;
if(pf<=1)
{
while (c1[i+j]==c2[j] && j<=s2.length())
{
j++;
System.out.println(s2.length()+" "+j);
if(j>=s2.length())
{
System.out.println("first match of(The) :->"+i);
}
pf=pf+1;
}
}
}
for(int i=0;i<(s3.length()-s6.length()+1);i++)
{
int j=0;
if(pl<=1)
{
while (c5[i+j]==c3[j] && j<=s6.length())
{
j++;
System.out.println(s6.length()+" "+j);
if(j>=s6.length())
{
System.out.println((s3.length()-i-3));
pl=pl+1;
}
}
}
}
}
}

//finding a particular word any where inthe string and printing its index and occurence
class IndOc
{
public static void main(String[] args)
{
String s="this is hyderabad city and this is";
System.out.println("the given string is ");
System.out.println("----------"+s);
char ch[]=s.toCharArray();
System.out.println(" ----word is found at ");
int j=0,noc=0;
for(int i=0;i<ch.length;i++)
{
j=i;
if(ch[i]=='i' && ch[j+1]=='s')
{
System.out.println(" index "+i);
noc++;
}
}
System.out.println("----- no of occurences are "+noc);
}
}

String match = "hello";
String text = "0123456789hello0123456789hello";
int j = 0;
String indxOfmatch = "";
for (int i = -1; i < text.length()+1; i++) {
j = text.indexOf("hello", i);
if (i>=j && j > -1) {
indxOfmatch += text.indexOf("hello", i)+" ";
}
}
System.out.println(indxOfmatch);

If you're going to scan for 'n' matches of the search string, I'd recommend using regular expressions.
They have a steep learning curve, but they'll save you hours when it comes to complex searches.

for multiple occurrence and the character found in string??yes or no
import java.io.BufferedReader;
import java.io.InputStreamReader;
public class SubStringtest {
public static void main(String[] args)throws Exception {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
System.out.println("enter the string");
String str=br.readLine();
System.out.println("enter the character which you want");
CharSequence ch=br.readLine();
boolean bool=str.contains(ch);
System.out.println("the character found is " +bool);
int position=str.indexOf(ch.toString());
while(position>=0){
System.out.println("the index no of character is " +position);
position=str.indexOf(ch.toString(),position+1);
}
}
}

public int NumberWordsInText(String FullText_, String WordToFind_, int[] positions_)
{
int iii1=0;
int iii2=0;
int iii3=0;
while((iii1=(FullText_.indexOf(WordToFind_,iii1)+1))>0){iii2=iii2+1;}
// iii2 is the number of the occurences
if(iii2>0) {
positions_ = new int[iii2];
while ((iii1 = (FullText_.indexOf(WordToFind_, iii1) + 1)) > 0) {
positions_[iii3] = iii1-1;
iii3 = iii3 + 1;
System.out.println("position=" + positions_[iii3 - 1]);
}
}
return iii2;
}

class Main{
public static int string(String str, String str1){
for (int i = 0; i <= str.length() - str1.length(); i++){
int j;
for (j = 0; j < str1.length(); j++) {
if (str1.charAt(j) != str.charAt(i + j)) {
break;
}
}
if (j == str1.length()) {
return i;
}}
return -1;
}
public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
System.out.println("Enter the string");
String str=sc.nextLine();
System.out.println("Enter the Substring");
String str1=sc.nextLine();
System.out.println("The position of the Substring is "+string(str, str1));
}
}