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
I'm currently trying to find a way to get my program to run more efficiently. If anyone has any ideas on how to optimize my program, i would appreciate it a lot.
Also, the output has to display in alphabetic order without any duplicates.
package Palindrome;
import java.io.*;
import java.util.*;
public class Palindrome {
public static void main(String[] args)throws IOException{
Scanner scan;
scan = new Scanner(System.in);
while (scan.hasNext()) {
subPal(scan.next());
}
if (scan != null) {
scan.close();
}}
private static void subPal(String str) {
String s1 = "";
int N = str.length();
ArrayList<String> palindromeArray = new ArrayList<String>();
for (int i = 2; i <= N; i++) {
for (int j = 0; j <= N; j++) {
int k = i + j - 1;
if (k >= N)
continue;
s1 = str.substring(j, i + j);
if (s1.equals(new StringBuilder(s1).reverse().toString())&& !palindromeArray.contains(s1)) {
palindromeArray.add(s1);
}
}
}
Collections.sort(palindromeArray);
for (String s : palindromeArray)
System.out.println(s );
}
}
This question already has answers here:
Printing reverse of any String without using any predefined function?
(34 answers)
Closed 8 years ago.
I was asked this in a technical interview. I have no idea whatsoever please please help me.
it goes in infinite loop. I just cant find the correct logic.
not once, but twice i came across this kind of a question, so please help
public static int numberOfCharsInString(String sentence)
{
int numberOfChars = 0,i=0;
while (!sentence.equals(""))
{
sentence = sentence.substring(1);
++numberOfChars;
}
return numberOfChars;
}
public static void reverseSequenceOfWords(String inp)
{
int len=numberOfCharsInString(inp);
char[] in=inp.toCharArray();
int i=0;
for(i=len-1;i>=0;i--)
{
if(in[i]==' ')
{
while(!in.equals("")||in.equals(" "))
{
System.out.print(in[i]+" ");
}
}
else if(in[i]=='\0')
{
break;
}
}
}
public static void main(String[] args)
{
int length=0;
String inpstring = "";
InputStreamReader input = new InputStreamReader(System.in);
BufferedReader reader = new BufferedReader(input);
try
{
System.out.print("Enter a string to reverse:");
inpstring = reader.readLine();
length=numberOfCharsInString(inpstring);
System.out.println("Number of Characters: "+length);
reverseSequenceOfWords(inpstring);
}
catch (Exception e)
{
e.printStackTrace();
}
}
String[] array = "Are you crazy".split(" ");
for (int i = array.length - 1; i >= 0; --i) {
System.out.print(array[i] + " ");
}
Brute forced this so hard lol
public static void main (String args[]){
String input = new Scanner(System.in).nextLine();
input+=" ";
ArrayList<String> words = new ArrayList<String>();
int start = 0;
for(int i=0; i<input.length(); i++){
if(input.charAt(i)==' '){
String toAdd="";
for(int r=start; r<i; r++){
toAdd+=input.charAt(r);
}
words.add(toAdd);
start = i+1;
}
}
for(int i=words.size()-1; i>=0; i--){
System.out.print(words.get(i)+" ");
}
}
I've used String.length() and String.substring()and String.charAt() - I hope that is allowed.
private static class Word {
private final String message;
private final int start;
private final int end;
public Word(String message, int start, int end) {
this.message = message;
this.start = start;
this.end = end;
}
#Override
public String toString() {
return message.substring(start, end);
}
}
private Word[] split(String message) {
// Split it into words - there cannot be more words than characters in the message.
int[] spaces = new int[message.length()];
// How many words.
int nWords = 0;
// Pretend there's a space at the start.
spaces[0] = -1;
// Walk the message.
for (int i = 0; i < message.length(); i++) {
if (message.charAt(i) == ' ') {
spaces[++nWords] = i;
}
}
// Record the final position.
spaces[++nWords] = message.length();
// Build the word array.
Word[] words = new Word[nWords];
for (int i = 0; i < nWords; i++) {
words[i] = new Word(message, spaces[i] + 1, spaces[i + 1]);
}
return words;
}
private String reverse(String message) {
Word[] split = split(message);
String reversed = "";
for ( int i = split.length - 1; i >= 0; i--) {
reversed += split[i].toString();
if ( i > 0 ) {
reversed += " ";
}
}
return reversed;
}
public void test() {
String message = "Hello how are you today?";
System.out.println(reverse(message));
}
prints
today? you are how Hello
Much more minimal but less useful. Only uses length, charAt and substring again:
public void printWordsReversed(String message) {
int end = message.length();
for ( int i = end - 1; i >= 0; i--) {
if ( message.charAt(i) == ' ') {
System.out.print(message.substring(i+1, end)+" ");
end = i;
}
}
System.out.print(message.substring(0, end));
}
The only function i'm still using is the IndexOf function, but that is not that hard to create for yourself.
static void Main(string[] args)
{
string sentence = "are you cracy";
int length = Program.StringLength(sentence);
int currentpos = 0;
List<string> wordList = new List<string>();
int wordCount = 0;
while (currentpos < length)
{
// find the next space
int spacepos = sentence.IndexOf(' ', currentpos);
string word;
if (spacepos < 0)
{
// end of string reached.
word = sentence.Substring(currentpos, length - currentpos);
wordList.Add(word);
wordCount++;
// no need to continue.
break;
}
word = sentence.Substring(currentpos, spacepos - currentpos);
wordList.Add(word);
wordCount++;
currentpos = spacepos + 1;
}
// display
for (int i = wordList.Count - 1; i >= 0; i--)
{
// after first word is display, add spaces to the output
if (i < wordList.Count - 1)
{
Console.WriteLine(" ");
}
// display word
Console.WriteLine(wordList[i]);
}
}
public static int StringLength(String sentence)
{
int numberOfChars = 0;
while (!sentence.Equals(""))
{
sentence = sentence.Substring(1);
++numberOfChars;
}
return numberOfChars;
}
In my program, the user enters a string, and it first finds the largest mode of characters in the string. Next, my program is supposed to remove all duplicates of a character in a string, (user input: aabc, program prints: abc) which I'm not entirely certain on how to do. I can get it to remove duplicates from some strings, but not all. For example, when the user puts "aabc" it will print "abc", but if the user puts "aabbhh", it will print "abbhh." Also, before I added the removeDup method to my program, it would only print the maxMode once, but after I added the removeDup method, it began to print the maxMode twice. How do I keep it from printing it twice?
Note: I cannot convert the strings to an array.
import java.util.Scanner;
public class JavaApplication3 {
static class MyStrings {
String s;
void setMyStrings(String str) {
s = str;
}
int getMode() {
int i;
int j;
int count = 0;
int maxMode = 0, maxCount = 1;
for (i = 0; i< s.length(); i++) {
maxCount = count;
count = 0;
for (j = s.length()-1; j >= 0; j--) {
if (s.charAt(j) == s.charAt(i))
count++;
if (count > maxCount){
maxCount = count;
maxMode = i;
}
}
}
System.out.println(s.charAt(maxMode)+" = largest mode");
return maxMode;
}
String removeDup() {
getMode();
int i;
int j;
String rdup = "";
for (i = 0; i< s.length(); i++) {
int count = 1;
for (j = 0; j < rdup.length(); j++) {
if (s.charAt(i) == s.charAt(j)){
count++;
}
}
if (count == 1){
rdup += s.charAt(i);
}
}
System.out.print(rdup);
System.out.println();
return rdup;
}
}
public static void main (String[] args) {
Scanner in = new Scanner(System.in);
MyStrings setS = new MyStrings();
String s;
System.out.print("Enter string:");
s = in.nextLine();
setS.setMyStrings(s);
setS.getMode();
setS.removeDup();
}
}
Try this method...should work fine!
String removeDup()
{
getMode();
int i;
int j;
String rdup = "";
for (i = 0; i< s.length(); i++) {
int count = 1;
for (j = i+1; j < s.length(); j++) {
if (s.charAt(i) == s.charAt(j)) {
count++;
}
}
if (count == 1){
rdup += s.charAt(i);
}
}
// System.out.print(rdup);
System.out.println();
return rdup;
}
Welcome to StackOverflow!
You're calling getMode() both outside and inside of removeDup(), which is why it's printing it twice.
In order to remove all duplicates, you'll have to call removeDup() over and over until all the duplicates are gone from your string. Right now you're only calling it once.
How might you do that? Think about how you're detecting duplicates, and use that as the end condition for a while loop or similar.
Happy coding!
Shouldn't this be an easier way? Also, i'm still learning.
import java.util.*;
public class First {
public static void main(String arg[])
{
Scanner sc= new Scanner(System.in);
StringBuilder s=new StringBuilder(sc.nextLine());
//String s=new String();
for(int i=0;i<s.length();i++){
String a=s.substring(i, i+1);
while(s.indexOf(a)!=s.lastIndexOf(a)){s.deleteCharAt(s.lastIndexOf(a));}
}
System.out.println(s.toString());
}
}
You can do this:
public static void main(String[] args) {
String str = new String("PINEAPPLE");
Set <Character> letters = new <Character>HashSet();
for (int i = 0; i < str.length(); i++) {
letters.add(str.charAt(i));
}
System.out.println(letters);
}
I think an optimized version which supports ASCII codes can be like this:
public static void main(String[] args) {
System.out.println(removeDups("*PqQpa abbBBaaAAzzK zUyz112235KKIIppP!!QpP^^*Www5W38".toCharArray()));
}
public static String removeDups(char []input){
long ocr1=0l,ocr2=0l,ocr3=0;
int index=0;
for(int i=0;i<input.length;i++){
int val=input[i]-(char)0;
long ocr=val<126?val<63?ocr1:ocr2:ocr3;
if((ocr& (1l<<val))==0){//not duplicate
input[index]=input[i];
index++;
}
if(val<63)
ocr1|=(1l<<val);
else if(val<126)
ocr2|=(1l<<val);
else
ocr3|=(1l<<val);
}
return new String(input,0,index);
}
please keep in mind that each of orc(s) represent a mapping of a range of ASCII characters and each java long variable can grow as big as (2^63) and since we have 128 characters in ASCII so we need three ocr(s) which basically maps the occurrences of the character to a long number.
ocr1: (char)0 to (char)62
ocr2: (char)63 to (char)125
ocr3: (char)126 to (char)128
Now if a duplicate was found the
(ocr& (1l<<val))
will be greater than zero and we skip that char and finally we can create a new string with the size of index which shows last non duplicate items index.
You can define more orc(s) and support other character-sets if you want.
Can use HashSet as well as normal for loops:
public class RemoveDupliBuffer
{
public static String checkDuplicateNoHash(String myStr)
{
if(myStr == null)
return null;
if(myStr.length() <= 1)
return myStr;
char[] myStrChar = myStr.toCharArray();
HashSet myHash = new HashSet(myStrChar.length);
myStr = "";
for(int i=0; i < myStrChar.length ; i++)
{
if(! myHash.add(myStrChar[i]))
{
}else{
myStr += myStrChar[i];
}
}
return myStr;
}
public static String checkDuplicateNo(String myStr)
{
// null check
if (myStr == null)
return null;
if (myStr.length() <= 1)
return myStr;
char[] myChar = myStr.toCharArray();
myStr = "";
int tail = 0;
int j = 0;
for (int i = 0; i < myChar.length; i++)
{
for (j = 0; j < tail; j++)
{
if (myChar[i] == myChar[j])
{
break;
}
}
if (j == tail)
{
myStr += myChar[i];
tail++;
}
}
return myStr;
}
public static void main(String[] args) {
String myStr = "This is your String";
myStr = checkDuplicateNo(myStr);
System.out.println(myStr);
}
Try this simple answer- works well for simple character string accepted as user input:
import java.util.Scanner;
public class string_duplicate_char {
String final_string = "";
public void inputString() {
//accept string input from user
Scanner user_input = new Scanner(System.in);
System.out.println("Enter a String to remove duplicate Characters : \t");
String input = user_input.next();
user_input.close();
//convert string to char array
char[] StringArray = input.toCharArray();
int StringArray_length = StringArray.length;
if (StringArray_length < 2) {
System.out.println("\nThe string with no duplicates is: "
+ StringArray[1] + "\n");
} else {
//iterate over all elements in the array
for (int i = 0; i < StringArray_length; i++) {
for (int j = i + 1; j < StringArray_length; j++) {
if (StringArray[i] == StringArray[j]) {
int temp = j;//set duplicate element index
//delete the duplicate element by copying the adjacent elements by one place
for (int k = temp; k < StringArray_length - 1; k++) {
StringArray[k] = StringArray[k + 1];
}
j++;
StringArray_length--;//reduce char array length
}
}
}
}
System.out.println("\nThe string with no duplicates is: \t");
//print the resultant string with no duplicates
for (int x = 0; x < StringArray_length; x++) {
String temp= new StringBuilder().append(StringArray[x]).toString();
final_string=final_string+temp;
}
System.out.println(final_string);
}
public static void main(String args[]) {
string_duplicate_char object = new string_duplicate_char();
object.inputString();
}
}
Another easy solution to clip the duplicate elements in a string using HashSet and ArrayList :
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Iterator;
import java.util.Scanner;
public class sample_work {
public static void main(String args[]) {
String input = "";
System.out.println("Enter string to remove duplicates: \t");
Scanner in = new Scanner(System.in);
input = in.next();
in.close();
ArrayList<Character> String_array = new ArrayList<Character>();
for (char element : input.toCharArray()) {
String_array.add(element);
}
HashSet<Character> charset = new HashSet<Character>();
int array_len = String_array.size();
System.out.println("\nLength of array = " + array_len);
if (String_array != null && array_len > 0) {
Iterator<Character> itr = String_array.iterator();
while (itr.hasNext()) {
Character c = (Character) itr.next();
if (charset.add(c)) {
} else {
itr.remove();
array_len--;
}
}
}
System.out.println("\nThe new string with no duplicates: \t");
for (int i = 0; i < array_len; i++) {
System.out.println(String_array.get(i).toString());
}
}
}
your can use this simple code and understand how to remove duplicates values from string.I think this is the simplest way to understand this problem.
class RemoveDup
{
static int l;
public String dup(String str)
{
l=str.length();
System.out.println("length"+l);
char[] c=str.toCharArray();
for(int i=0;i<l;i++)
{
for(int j=0;j<l;j++)
{
if(i!=j)
{
if(c[i]==c[j])
{
l--;
for(int k=j;k<l;k++)
{
c[k]=c[k+1];
}
j--;
}
}
}
}
System.out.println("after concatination lenght:"+l);
StringBuilder sd=new StringBuilder();
for(int i=0;i<l;i++)
{
sd.append(c[i]);
}
str=sd.toString();
return str;
}
public static void main(String[] ar)
{
RemoveDup obj=new RemoveDup();
Scanner sc=new Scanner(System.in);
String st,t;
System.out.println("enter name:");
st=sc.nextLine();
sc.close();
t=obj.dup(st);
System.out.println(t);
}
}
/*
* To change this license header, choose License Headers in Project Properties.
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package javaapplication26;
import java.util.*;
/**
*
* #author THENNARASU
*/
public class JavaApplication26 {
public static void main(String[] args) {
int i,j,k=0,count=0,m;
char a[]=new char[10];
char b[]=new char[10];
Scanner ob=new Scanner(System.in);
String str;
str=ob.next();
a=str.toCharArray();
int c=str.length();
for(j=0;j<c;j++)
{
for(i=0;i<j;i++)
{
if(a[i]==a[j])
{
count=1;
}
}
if(count==0)
{
b[k++]=a[i];
}
count=0;
}
for(m=0;b[m]!='\0';m++)
{
System.out.println(b[m]);
}
}
}
i wrote this program. Am using 2 char arrays instead. You can define the number of duplicate chars you want to eliminate from the original string and also shows the number of occurances of each character in the string.
public String removeMultipleOcuranceOfChar(String string, int numberOfChars){
char[] word1 = string.toCharArray();
char[] word2 = string.toCharArray();
int count=0;
StringBuilder builderNoDups = new StringBuilder();
StringBuilder builderDups = new StringBuilder();
for(char x: word1){
for(char y : word2){
if (x==y){
count++;
}//end if
}//end inner loop
System.out.println(x + " occurance: " + count );
if (count ==numberOfChars){
builderNoDups.append(x);
}else{
builderDups.append(x);
}//end if else
count = 0;
}//end outer loop
return String.format("Number of identical chars to be in or out of input string: "
+ "%d\nOriginal word: %s\nWith only %d identical chars: %s\n"
+ "without %d identical chars: %s",
numberOfChars,string,numberOfChars, builderNoDups.toString(),numberOfChars,builderDups.toString());
}
Try this simple solution for REMOVING DUPLICATE CHARACTERS/LETTERS FROM GIVEN STRING
import java.util.Scanner;
public class RemoveDuplicateLetters {
public static void main(String[] args) {
Scanner scn=new Scanner(System.in);
System.out.println("enter a String:");
String s=scn.nextLine();
String ans="";
while(s.length()>0)
{
char ch = s.charAt(0);
ans+= ch;
s = s.replace(ch+"",""); //Replacing all occurrence of the current character by a spaces
}
System.out.println("after removing all duplicate letters:"+ans);
}
}
In Java 8 we can do that using
private void removeduplicatecharactersfromstring() {
String myString = "aabcd eeffff ghjkjkl";
StringBuilder builder = new StringBuilder();
Arrays.asList(myString.split(" "))
.forEach(s -> {
builder.append(Stream.of(s.split(""))
.distinct().collect(Collectors.joining()).concat(" "));
});
System.out.println(builder); // abcd ef ghjkl
}
Given a string how can i figure out the number of times each char in a string repeats itself
ex: aaaabbaaDD
output: 4a2b2a2D
public static void Calc() {
Input();
int count = 1;
String compressed = "";
for (int i = 0; i < input.length(); i++) {
if (lastChar == input.charAt(i)) {
count++;
compressed += Integer.toString(count) + input.charAt(i);
}
else {
lastChar = input.charAt(i);
count = 1;
}
}
System.out.println(compressed);
}
What you'r looking for is "Run-length encoding". Here is the working code to do that;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class RunLengthEncoding {
public static String encode(String source) {
StringBuffer dest = new StringBuffer();
// iterate through input string
// Iterate the string N no.of.times where N is size of the string to find run length for each character
for (int i = 0; i < source.length(); i++) {
// By default run Length for all character is one
int runLength = 1;
// Loop condition will break when it finds next character is different from previous character.
while (i+1 < source.length() && source.charAt(i) == source.charAt(i+1)) {
runLength++;
i++;
}
dest.append(runLength);
dest.append(source.charAt(i));
}
return dest.toString();
}
public static String decode(String source) {
StringBuffer dest = new StringBuffer();
Pattern pattern = Pattern.compile("[0-9]+|[a-zA-Z]");
Matcher matcher = pattern.matcher(source);
while (matcher.find()) {
int number = Integer.parseInt(matcher.group());
matcher.find();
while (number-- != 0) {
dest.append(matcher.group());
}
}
return dest.toString();
}
public static void main(String[] args) {
String example = "WWWWWWWWWWWWBWWWWWWWWWWWWBBBWWWWWWWWWWWWWWWWWWWWWWWWBWWWWWWWWWWWWWW";
System.out.println(encode(example));
System.out.println(decode("1W1B1W1B1W1B1W1B1W1B1W1B1W1B"));
}
}
This program first finds the unique characters or numbers in a string. It will then check the frequency of occurance.
This program considers capital and small case as different characters. You can modify it if required by using ignorecase method.
import java.io.*;
public class RunLength {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
public static void main(String[] args) throws IOException {
System.out.println("Please enter the string");
String str = br.readLine();//the input string is in str
calculateFrequency(str);
}
private static void calculateFrequency(String str) {
int length = str.length();
String characters[] = new String[length];//to store all unique characters in string
int frequency[] = new int[length];//to store the frequency of the characters
for (int i = 0; i < length; i++) {
characters[i] = null;
frequency[i] = 0;
}
//To get unique characters
char temp;
String temporary;
int uniqueCount = 0;
for (int i = 0; i < length; i++) {
int flag = 0;
temp = str.charAt(i);
temporary = "" + temp;
for (int j = 0; j < length; j++) {
if (characters[j] != null && characters[j].equals(temporary)) {
flag = 1;
break;
}
}
if (flag == 0) {
characters[uniqueCount] = temporary;
uniqueCount++;
}
}
// To get the frequency of the characters
for(int i=0;i<length;i++){
temp=str.charAt(i);
temporary = ""+temp;
for(int j=0;i<characters.length;j++){
if(characters[j].equals(temporary)){
frequency[j]++;
break;
}
}
}
// To display the output
for (int i = 0; i < length; i++) {
if (characters[i] != null) {
System.out.println(characters[i]+" "+frequency[i]);
}
}
}}
Some hints: In your code sample you also need to reset count to 0 when the run ends (when you update lastChar). And you need to output the final run (after the loop is done). And you need some kind of else or continue between the two cases.
#Balarmurugan k's solution is better - but just by improving upon your code I came up with this -
String input = "aaaabbaaDD";
int count = 0;
char lastChar = 0;
int inputSize = input.length();
String output = "";
for (int i = 0; i < inputSize; i++) {
if (i == 0) {
lastChar = input.charAt(i);
count++;
} else {
if (lastChar == input.charAt(i)) {
count++;
} else {
output = output + count + "" + lastChar;
count = 1;
lastChar = input.charAt(i);
}
}
}
output = output + count + "" + lastChar;
System.out.println(output);