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Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: Index out of bounds

I was getting this error while running the following code. I couldn't find out what was wrong with the code.
As far as I can see, there's some issue with my second character array. But couldn't find out what was wrong. First tried running the last loop before temp_count. Then also tried temp_count±1. Yet, I failed. I have also tried taking different array size. still no luck
import java.util.Scanner;
public class oop2
{
public static void main(String[] args)
{
Scanner sc = new Scanner(System.in);
String str = new String();
int temp_count = 0;
//New input of string
str=sc.nextLine();
char[] c = str.toCharArray();
char[] temp = new char[temp_count];
//Converting uppercase to lower case for convenience
for(int i=0; i<str.length(); i++)
{
if (Character.isUpperCase(c[i]))
{
c[i]=(char) (c[i]+32);
}
}
//verifying whether the alphabet exists
for(char x = 'a'; x<='z'; x++)
{
int count=0;
for(int i=0; c[i]!='\0'; i++)
{
if (c[i]==x)
{
count++;
}
}
//if the alphabet is not found, then putting the alphabet in
if (count==0)
{
temp[temp_count]=x;
temp_count++;
}
}
//Verifying whether it's a pangram or not
if (temp_count==0)
{
System.out.println("Pangram");
}
else
{
//if not pangram then this part will execute
System.out.println("Not Pangram");
System.out.printf("Missing Characters: ");
//printing out the missing character
for(int i=0; i<temp_count-1; i++)
{
System.out.print(temp[i]+", ");
}
}
sc.close();
}
}

import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String str = new String();
int temp_count = 0;
//New input of string
str = sc.nextLine();
char[] c = str.toCharArray();
char[] temp = new char[0];
//Converting uppercase to lower case for convenience
for (int i = 0; i < str.length(); i++) {
if (Character.isUpperCase(c[i])) {
c[i] = (char)(c[i] + 32);
}
}
//verifying whether the alphabet exists
for (char x = 'a'; x <= 'z'; x++) {
int count = 0;
for (int i = 0; c[i] != '\0'; i++) {
if (c[i] == x) {
count++;
}
}
//if the alphabet is not found, then putting the alphabet in
if (count == 0) {
char [] copy = new char[temp.length+1];
for (int i = 0; i < temp.length; i++) {
copy[i]=temp[i];
}
copy[copy.length-1] = x;
temp=copy;
}
}
//Verifying whether it's a pangram or not
if (temp_count == 0) {
System.out.println("Pangram");
} else {
//if not pangram then this part will execute
System.out.println("Not Pangram");
System.out.printf("Missing Characters: ");
//printing out the missing character
for (int i = 0; i < temp_count - 1; i++) {
System.out.print(temp[i] + ", ");
}
}
sc.close();
}
}
Blockquote

Why is there no output getting displayed? Here I am trying to display two strings in alternate positions

In the below code I am trying to print two strings in a single char array in alternate positions, where second string should be stored in the reverse order.
For example: str1="happy" str2= "sadly" arr="hyalpdpays", where both strings should be of the same size.
import java.util.*;
//*class Main declaration done here*
class Main{
public static void main (String args[])
{
char[] arr=new char[100];
int flen=0;
Scanner sc=new Scanner(System.in);
String str1=sc.nextLine();
String str2=sc.nextLine();
if(str1.length()==str2.length())
{
flen = str1.length()+str2.length();
for(int i=0, j=flen, k=0; i<(flen/2) && j>=0 && k<flen; i=i+2 , j=j-2, k++)
{
arr[k]=str1.charAt(i);
k=k+1;
arr[k]=str2.charAt(i);
}
}
for(int i=0; i<flen; i++)
{
System.out.println(arr[i]);
}
}
}

I did some changes to your code this works fine with me
if(str1.length()==str2.length())
{
flen = str1.length()+str2.length();
int lastIndex = (str2.length()-1);
for(int i=0, j=lastIndex, k=0; i<flen; j--, k++,i++)
{
arr[i]=str1.charAt(k);
i++;
arr[i]=str2.charAt(j);
}
}
for(int i=0; i<flen; i++)
{
System.out.print(arr[i]);
}
so it would start in the index 0 of string 1 then the last index of string 2 and so on. the output will be as u mentioned "hyalpdpays".

You can do it simply as follows:
public class Main {
public static void main(String[] args) {
String str1 = "happy";
String str2 = "sadly";
StringBuilder sb = new StringBuilder();
if (str1.length() == str2.length()) {
for (int i = 0; i < str1.length(); i++) {
sb.append(str1.charAt(i));// Chars of str1 from beginning
sb.append(str2.charAt(str2.length() - i - 1));// Chars of str2 from end
}
} else {
System.out.println("Strings are of different lengths");
}
System.out.println(sb);
}
}
Output:
hyalpdpays

How to count capital letters with compareTo() method?

Is there a way to count capital letters in a string using the method compareTo()? This is my current code so far, I don't know what to add in the if statement.
import java.util.*;
public class countcapitalletters
{
public static void main(String args[])
{
Scanner scan = new Scanner(System.in);
System.out.print("enter a string");
String input = scan.nextLine();
int count = 0;
for(int i=0; i<=input.length()-1; i++)
{
if(input.substring(i,i+1)
{
count = count+1;
}
}
System.out.println(count);
}
}
i dont know what to add to my if statement.

You can write a method as follows:
public int countUpperCase(String input) {
int count = 0;
for (int i = 0; i < input.length(); i++) {
String currentChar = input.substring(i, i + 1);
if (currentChar.compareTo("A") >= 0 && currentChar.compareTo("Z") <= 0) {
count = count + 1;
}
}
return count;
}

Remove repeated characters in a string

I need to write a static method that takes a String as a parameter and returns a new String obtained by replacing every instance of repeated adjacent letters with a single instance of that letter without using regular expressions. For example if I enter "maaaakkee" as a String, it returns "make".
I already tried the following code, but it doesn't seem to display the last character.
Here's my code:
import java.util.Scanner;
public class undouble {
public static void main(String [] args){
Scanner console = new Scanner(System.in);
System.out.println("enter String: ");
String str = console.nextLine();
System.out.println(removeSpaces(str));
}
public static String removeSpaces(String str){
String ourString="";
int j = 0;
for (int i=0; i<str.length()-1 ; i++){
j = i+1;
if(str.charAt(i)!=str.charAt(j)){
ourString+=str.charAt(i);
}
}
return ourString;
}
}

You could use regular expressions for that.
For instance:
String input = "ddooooonnneeeeee";
System.out.println(input.replaceAll("(.)\\1{1,}", "$1"));
Output:
done
Pattern explanation:
"(.)\\1{1,}" means any character (added to group 1) followed by itself at least once
"$1" references contents of group 1

maybe:
for (int i=1; i<str.length() ; i++){
j = i+1;
if(str.charAt(i)!=str.charAt(j)){
ourString+=str.charAt(i);
}
}

The problem is with your condition. You say compare i and i+1 in each iteration and in last iteration you have both i and j pointing to same location so it will never print the last character. Try this unleass you want to use regex to achive this:
EDIT:
public void removeSpaces(String str){
String ourString="";
for (int i=0; i<str.length()-1 ; i++){
if(i==0){
ourString = ""+str.charAt(i);
}else{
if(str.charAt(i-1) != str.charAt(i)){
ourString = ourString +str.charAt(i);
}
}
}
System.out.println(ourString);
}

if you cannot use replace or replaceAll, here is an alternative. O(2n), O(N) for stockage and O(N) for creating the string. It removes all repeated chars in the string put them in a stringbuilder.
input : abcdef , output : abcdef
input : aabbcdeef, output : cdf
private static String remove_repeated_char(String str)
{
StringBuilder result = new StringBuilder();
HashMap<Character, Integer> items = new HashMap<>();
for (int i = 0; i < str.length(); i++)
{
Character current = str.charAt(i);
Integer ocurrence = items.get(current);
if (ocurrence == null)
items.put(current, 1);
else
items.put(current, ocurrence + 1);
}
for (int i = 0; i < str.length(); i++)
{
Character current = str.charAt(i);
Integer ocurrence = items.get(current);
if (ocurrence == 1)
result.append(current);
}
return result.toString();
}

import java.util.*;
public class string2 {
public static void main(String[] args) {
//removes repeat character from array
Scanner sc=new Scanner(System.in);
StringBuffer sf=new StringBuffer();
System.out.println("enter a string");
sf.append(sc.nextLine());
System.out.println("string="+sf);
int i=0;
while( i<sf.length())
{
int j=1+i;
while(j<sf.length())
{
if(sf.charAt(i)==sf.charAt(j))
{
sf.deleteCharAt(j);
}
else
{
j=j+1;
}
}
i=i+1;
}
System.out.println("string="+sf);
}
}

Input AABBBccDDD, Output BD
Input ABBCDDA, Outout C
private String reducedString(String s){
char[] arr = s.toCharArray();
String newString = "";
Map<Character,Integer> map = new HashMap<Character,Integer>();
map.put(arr[0],1);
for(int index=1;index<s.length();index++)
{
Character key = arr[index];
int value;
if(map.get(key) ==null)
{
value =0;
}
else
{
value = map.get(key);
}
value = value+1;
map.put(key,value);
}
Set<Character> keyset = map.keySet();
for(Character c: keyset)
{
int value = map.get(c);
if(value%2 !=0)
{
newString+=c;
}
}
newString = newString.equals("")?"Empty String":newString;
return newString;
}

public class RemoveDuplicateCharecterInString {
static String input = new String("abbbbbbbbbbbbbbbbccccd");
static String output = "";
public static void main(String[] args)
{
// TODO Auto-generated method stub
for (int i = 0; i < input.length(); i++) {
char temp = input.charAt(i);
boolean check = false;
for (int j = 0; j < output.length(); j++) {
if (output.charAt(j) == input.charAt(i)) {
check = true;
}
}
if (!check) {
output = output + input.charAt(i);
}
}
System.out.println(" " + output);
}
}
Answer : abcd

public class RepeatedChar {
public static void main(String[] args) {
String rS = "maaaakkee";
String outCome= rS.charAt(0)+"";
int count =0;
char [] cA =rS.toCharArray();
for(int i =0; i+1<cA.length; ++i) {
if(rS.charAt(i) != rS.charAt(i+1)) {
outCome += rS.charAt(i+1);
}
}
System.out.println(outCome);
}
}

TO WRITE JAVA PROGRAM TO REMOVE REPEATED CHARACTERS:
package replace;
public class removingrepeatedcharacters
{
public static void main(String...args){
int i,j=0,count=0;
String str="noordeen";
String str2="noordeen";
char[] ch=str.toCharArray();
for(i=0;i<=5;i++)
{
count=0;
for(j=0;j<str2.length();j++)
{
if(ch[i]==str2.charAt(j))
{
count++;
System.out.println("at the index "+j +"position "+ch[i]+ "+ count is"+count);
if(count>=2){
str=str2;
str2=str.replaceFirst(Character.toString(ch[j]),Character.toString(' '));
}
System.out.println("after replacing " +str2);
}
}
}
}
}

String outstr = "";
String outstring = "";
for(int i = 0; i < str.length() - 1; i++) {
if(str.charAt(i) != str.charAt(i + 1)) {
outstr = outstr + str.charAt(i);
}
outstring = outstr + str.charAt(i);
}
System.out.println(outstring);

public static void remove_duplicates(String str){
String outstr="";
String outstring="";
for(int i=0;i<str.length()-1;i++) {
if(str.charAt(i)!=str.charAt(i+1)) {
outstr=outstr+str.charAt(i);
}
outstring=outstr+str.charAt(i);
}
System.out.println(outstring);
}

More fun with java 7:
System.out.println("11223344445555".replaceAll("(?<nums>.+)\\k<nums>+","${nums}"));
No more cryptic numbers in regexes.

public static String removeDuplicates(String str) {
String str2 = "" + str.charAt(0);
for (int i = 1; i < str.length(); i++) {
if (str.charAt(i - 1) == str.charAt(i) && i != 0) {
continue;
}
str2 = str2 + str.charAt(i);
}
return str2;
}

permutations of a string using iteration

I'm trying to find permutation of a given string, but I want to use iteration. The recursive solution I found online and I do understand it, but converting it to an iterative solution is really not working out. Below I have attached my code. I would really appreciate the help:
public static void combString(String s) {
char[] a = new char[s.length()];
//String temp = "";
for(int i = 0; i < s.length(); i++) {
a[i] = s.charAt(i);
}
for(int i = 0; i < s.length(); i++) {
String temp = "" + a[i];
for(int j = 0; j < s.length();j++) {
//int k = j;
if(i != j) {
System.out.println(j);
temp += s.substring(0,j) + s.substring(j+1,s.length());
}
}
System.out.println(temp);
}
}

Following up on my related question comment, here's a Java implementation that does what you want using the Counting QuickPerm Algorithm:
public static void combString(String s) {
// Print initial string, as only the alterations will be printed later
System.out.println(s);
char[] a = s.toCharArray();
int n = a.length;
int[] p = new int[n]; // Weight index control array initially all zeros. Of course, same size of the char array.
int i = 1; //Upper bound index. i.e: if string is "abc" then index i could be at "c"
while (i < n) {
if (p[i] < i) { //if the weight index is bigger or the same it means that we have already switched between these i,j (one iteration before).
int j = ((i % 2) == 0) ? 0 : p[i];//Lower bound index. i.e: if string is "abc" then j index will always be 0.
swap(a, i, j);
// Print current
System.out.println(join(a));
p[i]++; //Adding 1 to the specific weight that relates to the char array.
i = 1; //if i was 2 (for example), after the swap we now need to swap for i=1
}
else {
p[i] = 0;//Weight index will be zero because one iteration before, it was 1 (for example) to indicate that char array a[i] swapped.
i++;//i index will have the option to go forward in the char array for "longer swaps"
}
}
}
private static String join(char[] a) {
StringBuilder builder = new StringBuilder();
builder.append(a);
return builder.toString();
}
private static void swap(char[] a, int i, int j) {
char temp = a[i];
a[i] = a[j];
a[j] = temp;
}

List<String> results = new ArrayList<String>();
String test_str = "abcd";
char[] chars = test_str.toCharArray();
results.add(new String("" + chars[0]));
for(int j=1; j<chars.length; j++) {
char c = chars[j];
int cur_size = results.size();
//create new permutations combing char 'c' with each of the existing permutations
for(int i=cur_size-1; i>=0; i--) {
String str = results.remove(i);
for(int l=0; l<=str.length(); l++) {
results.add(str.substring(0,l) + c + str.substring(l));
}
}
}
System.out.println("Number of Permutations: " + results.size());
System.out.println(results);
Example:
if we have 3 character string e.g. "abc", we can form permuations as below.
1) construct a string with first character e.g. 'a' and store that in results.
char[] chars = test_str.toCharArray();
results.add(new String("" + chars[0]));
2) Now take next character in string (i.e. 'b') and insert that in all possible positions of previously contsructed strings in results. Since we have only one string in results ("a") at this point, doing so gives us 2 new strings 'ba', 'ab'. Insert these newly constructed strings in results and remove "a".
for(int i=cur_size-1; i>=0; i--) {
String str = results.remove(i);
for(int l=0; l<=str.length(); l++) {
results.add(str.substring(0,l) + c + str.substring(l));
}
}
3) Repeat 2) for every character in the given string.
for(int j=1; j<chars.length; j++) {
char c = chars[j];
....
....
}
This gives us "cba", "bca", "bac" from "ba" and "cab", "acb" and "abc" from "ab"

Work queue allows us to create an elegant iterative solution for this problem.
static List<String> permutations(String string) {
List<String> permutations = new LinkedList<>();
Deque<WorkUnit> workQueue = new LinkedList<>();
// We need to permutate the whole string and haven't done anything yet.
workQueue.add(new WorkUnit(string, ""));
while (!workQueue.isEmpty()) { // Do we still have any work?
WorkUnit work = workQueue.poll();
// Permutate each character.
for (int i = 0; i < work.todo.length(); i++) {
String permutation = work.done + work.todo.charAt(i);
// Did we already build a complete permutation?
if (permutation.length() == string.length()) {
permutations.add(permutation);
} else {
// Otherwise what characters are left?
String stillTodo = work.todo.substring(0, i) + work.todo.substring(i + 1);
workQueue.add(new WorkUnit(stillTodo, permutation));
}
}
}
return permutations;
}
A helper class to hold partial results is very simple.
/**
* Immutable unit of work
*/
class WorkUnit {
final String todo;
final String done;
WorkUnit(String todo, String done) {
this.todo = todo;
this.done = done;
}
}
You can test the above piece of code by wrapping them in this class.
import java.util.*;
public class AllPermutations {
public static void main(String... args) {
String str = args[0];
System.out.println(permutations(str));
}
static List<String> permutations(String string) {
...
}
}
class WorkUnit {
...
}
Try it by compiling and running.
$ javac AllPermutations.java; java AllPermutations abcd
The below implementation can also be easily tweaked to return a list of permutations in reverse order by using a LIFO stack of work instead of a FIFO queue.

import java.util.List;
import java.util.Set;
import java.util.ArrayList;
import java.util.HashSet;
public class Anagrams{
public static void main(String[] args)
{
String inpString = "abcd";
Set<String> combs = getAllCombs(inpString);
for(String comb : combs)
{
System.out.println(comb);
}
}
private static Set<String> getAllCombs(String inpString)
{
Set<String> combs = new HashSet<String>();
if( inpString == null | inpString.isEmpty())
return combs;
combs.add(inpString.substring(0,1));
Set<String> tempCombs = new HashSet<String>();
for(char a : inpString.substring(1).toCharArray())
{
tempCombs.clear();
tempCombs.addAll(combs);
combs.clear();
for(String comb : tempCombs)
{
combs.addAll(getCombs(comb,a));
}
}
return combs;
}
private static Set<String> getCombs(String comb, char a) {
Set<String> combs = new HashSet<String>();
for(int i = 0 ; i <= comb.length(); i++)
{
String temp = comb.substring(0, i) + a + comb.substring(i);
combs.add(temp);
//System.out.println(temp);
}
return combs;
}
}

Just posting my approach to the problem:
import java.util.ArrayDeque;
import java.util.Queue;
public class PermutationIterative {
public static void main(String[] args) {
permutationIterative("abcd");
}
private static void permutationIterative(String str) {
Queue<String> currentQueue = null;
int charNumber = 1;
for (char c : str.toCharArray()) {
if (currentQueue == null) {
currentQueue = new ArrayDeque<>(1);
currentQueue.add(String.valueOf(c));
} else {
int currentQueueSize = currentQueue.size();
int numElements = currentQueueSize * charNumber;
Queue<String> nextQueue = new ArrayDeque<>(numElements);
for (int i = 0; i < currentQueueSize; i++) {
String tempString = currentQueue.remove();
for (int j = 0; j < charNumber; j++) {
int n = tempString.length();
nextQueue.add(tempString.substring(0, j) + c + tempString.substring(j, n));
}
}
currentQueue = nextQueue;
}
charNumber++;
}
System.out.println(currentQueue);
}
}

package vishal villa;
import java.util.Scanner;
public class Permutation {
static void result( String st, String ans)
{
if(st.length() == 0)
System.out.println(ans +" ");
for(int i = 0; i<st.length(); i++)
{
char ch = st.charAt(i);
String r = st.substring(0, i) + st.substring(i + 1);
result(r, ans + ch);
}
}
public static void main(String[] args)
{
Scanner Sc = new Scanner(System.in);
System.out.println("enter the string");
String st = Sc.nextLine();
Permutation p = new Permutation();
p.result(st,"" );
}
}

// Java program to print all permutations of a
// given string.
public class Permutation
{
public static void main(String[] args)
{
String str = "ABC";
int n = str.length();
Permutation permutation = new Permutation();
permutation.permute(str, 0, n-1);
}
/**
* permutation function
* #param str string to calculate permutation for
* #param s starting index
* #param e end index
*/
private void permute(String str, int s, int e)
{
if (s == e)
System.out.println(str);
else
{
for (int i = s; i <= s; i++)
{
str = swap(str,l,i);
permute(str, s+1, e);
str = swap(str,l,i);
}
}
}
/**
* Swap Characters at position
* #param a string value
* #param i position 1
* #param j position 2
* #return swapped string
*/
public String swap(String a, int i, int j)
{
char temp;
char[] charArray = a.toCharArray();
temp = charArray[i] ;
charArray[i] = charArray[j];
charArray[j] = temp;
return String.valueOf(charArray);
}
}