String errors in convertor - java

so my code works for some words but not others it sprints out String index out of range: -1
im suppose to print out ub before every vowel and vowel cluster
ex dubious would be dubububious or cat loveo would be lubovudeo
String sentance, set;
sentance = "toster iooppp";
set= translate(sentance);
System.out.println(set);
}
public static String translate (String sentence){
String set = " ";
sentence= sentence.toLowerCase();
scan = new Scanner (sentence);
while (scan.hasNext()) {
set+= toUbbi (scan.next());
set += " ";
}
return set;
}
private static String toUbbi(String word ) {
String str= word;
String new_str="";
for (int i = 0; i < str.length(); i++) {
char c = str.charAt(i);
if (isVowel(c) && isVowel(str.charAt(i -1)) )
{ // If is a vowel
new_str += "ub" ;
}
new_str += c;
}
return new_str;
}
private static boolean isVowel(char c)
{
if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' ){
return true;}
return false;


in first iteration it gives error..
if (isVowel(c) && isVowel(str.charAt(i -1)) )
{ // If is a vowel
new_str += "ub" ;
}
so change like this..
if (isVowel(c))
{ // If is a vowel
new_str += "ub" ;
}
or change your loop..
for (int i = 1; i < str.length(); i++) {
// yur remain code...
}

Related

String Index Exception

I was trying to convert an expression from infix form to postfix form.I used String a, p , s for stack, postfix result expression, input expression respectively.
Every time I am getting this error:
Exception in thread "main" java.lang.StringIndexOutOfBoundsException:
String index out of range: -1 at
java.lang.String.charAt(String.java:658) at
javaapplication4.A.conversion(A.java:50) at
javaapplication4.A.main(A.java:83)
Please help me how can I solve it.
Here is my code:
import java.util.Scanner;
public class A {
String a="(", s = "", p = "";
int i, n = 1, top = 0, pp = 0;
void push(char ch) {
a = a + ch;
top = n;
n++;
}
void pop() {
n--;
top--;
}
int prio(char ch) {
int f = -1;
if (ch == '(') {
f = 0;
} else if (ch == '+' || ch == '-') {
f = 1;
} else if (ch == '*' || ch == '/' || ch == '%') {
f = 2;
} else if (ch == '^') {
f = 3;
}
return f;
}
void conversion() {
System.out.print("Enter infix form: ");
Scanner sd = new Scanner(System.in);
s = sd.nextLine();
//System.out.println(s);
int t, j, sz;
sz = s.length();
for (i = 0; i < sz; i++) {
if (s.charAt(i) >= '0' && s.charAt(i) <= '9') {
p = p + s.charAt(i);
pp++;
} else if (s.charAt(i) == '(') {
push('(');
} else if (s.charAt(i) == '-' || s.charAt(i) == '+' || s.charAt(i) == '*' || s.charAt(i) == '/' || s.charAt(i) == '%' || s.charAt(i) == '^') {
j = prio(s.charAt(i));
t = prio(a.charAt(top));
//System.out.println(t+" "+j);
while (j <= t) {
p = p + a.charAt(top);
pp++;
pop();
t = prio(a.charAt(top));
}
push(s.charAt(i));
} else if (s.charAt(i) == ')') {
while (a.charAt(top) != '(') {
p = p + a.charAt(top);
pp++;
pop();
}
pop();
}
}
while (a.charAt(top) != '(') {
p = p + a.charAt(top);
pp++;
pop();
}
pop();
}
void postfix() {
System.out.print("postfix form is: ");
System.out.println(p);
}
public static void main(String args[]) {
A h = new A();
h.conversion();
h.postfix();
//System.out.println(h.a);
//System.out.println(h.s);
}
}

Can you confirm if the error is here?
while (a.charAt(top) != '(')
Inside the loop you continuously pop(), which decrements from top, which has the risk of going negative if a match is never found. Even if the error is not here, it should check for that condition.

You may have made an extra pop() definition. Can you check this?

i am trying to take user input and display with a condition that it should be Alphabet

public static void main(String[] args) {
String name;
System.out.println("Please Enter Name");
Scanner s=new Scanner(System.in);
//s.nextLine();
name=s.nextLine();
if(isAlpha(name)) {
System.out.println("Name is: " +name);
}else {
System.out.println("please Enter valid name");
}
}
public static boolean isAlpha(String name) {
char[] charArray = name.toCharArray();
for (int i = 0; i < charArray.length; i++) {
char ch = charArray[i];
if (!(ch >= 'a' && ch <= 'z')) {
return false;
}
else if (!(ch >= 'A' && ch <= 'Z')) {
return false;
}
else {
return true;
}
}
return true;
}
i am trying to take user input and display it only if it satisfy a condition that it should contain alphabets only but it shows every entry as invalid why?

You can't have a character that is both uppercase & lowercase so one of your if blocks will always return false in the isAlpha method. Instead you could do
public static boolean isAlpha(String name) {
char[] charArray = name.toLowerCase().toCharArray();
for (int i = 0; i < charArray.length; i++) {
char ch = charArray[i];
if (!(ch >= 'a' && ch <= 'z')) {
return false;
}
}
return true;
}

Use this
public static boolean isAlpha(String s) {
return s != null && s.matches("^[a-zA-Z]*$");
}

Print the number of unique vowels in a string, Java

I need to find the number of distinct vowels. I came up with the code below but it can't make distinction between same vowels:
public static int count_Vowels(String str) {
str = str.toLowerCase();
int count = 0;
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == 'a' || str.charAt(i) == 'e' || str.charAt(i) == 'i'
|| str.charAt(i) == 'o' || str.charAt(i) == 'u') {
count++;
}
}
return count;
}

I would start with five variables (one for each vowel) set to 0, iterate the characters in the input and set the corresponding variable to 1 if I find a match, and simply return the accumulated value of said variables. Like,
public static int count_Vowels(String str) {
int a = 0, e = 0, i = 0, o = 0, u = 0;
for (char ch : str.toLowerCase().toCharArray()) {
if (ch == 'a') {
a = 1;
} else if (ch == 'e') {
e = 1;
} else if (ch == 'i') {
i = 1;
} else if (ch == 'o') {
o = 1;
} else if (ch == 'u') {
u = 1;
}
}
return a + e + i + o + u;
}

You could use Set data structure and instead of incrementing the counter just add vowels to the set. At the end you can return just the size of the set.

The problem in your code is that you are not counting the distinct vowels, but all the vowels in the string. A Java-8 way to this:
public static int countDistinctVowels(String str) {
str = str.toLowerCase();
int count = (int) str.chars() // get IntStream of chars
.mapToObj(c -> (char) c) // cast to char
.filter(c -> "aeiou".indexOf(c) > -1) // remove all non-vowels
.distinct() // keep the distinct values
.count(); // count the values
return count;
}
Also use proper Java naming conventions: countDistinctVowels, no count_Distinct_Vowels.

there's definitely an issue here with this counting. at the very least. you should rethink this:
if (str.charAt(i) == 'a' || str.charAt(i) == 'e' || str.charAt(i) == 'i'
|| str.charAt(i) == 'o' || str.charAt(i) == 'u')
count++;

You could use the method contains
public static int count_Vowels(String str) {
str = str.toLowerCase();
int count = 0;
count += string.contains("a") ? 1 : 0;
count += string.contains("e") ? 1 : 0;
count += string.contains("i") ? 1 : 0;
count += string.contains("o") ? 1 : 0;
count += string.contains("u") ? 1 : 0;
return count;
}

I made explanations in the comments to the code:
public static int count_Vowels(String str) {
str = str.toLowerCase();
Set<Character> setOfUsedChars = new HashSet<>(); // Here you store used vowels
for (int i = 0; i < str.length(); i++) {
if (str.charAt(i) == 'a' || str.charAt(i) == 'e' || str.charAt(i) == 'i'
|| str.charAt(i) == 'o' || str.charAt(i) == 'u') { // if currently checked character is vowel...
setOfUsedChars.add(str.charAt(i)); // add this vowel to setOfUsedChars
}
}
return setOfUsedChars.size(); // size of this sets is a number of vowels present in input String
}

static void vow(String input){
String output=input.toLowerCase();
int flaga=0,flage=0,flagi=0,flago=0,flagu=0;
for(int i=0;i<input.length();i++) {
if((output.charAt(i))=='a' && flaga==0) {
System.out.print(input.charAt(i)+" ");
flaga++;
}
if(output.charAt(i)=='e' && flage==0) {
System.out.print(input.charAt(i)+" ");
flage++;
}
if(output.charAt(i)=='i' && flagi==0) {
System.out.print(input.charAt(i)+" ");
flagi++;
}
if(output.charAt(i)=='o' && flago==0) {
System.out.print(input.charAt(i)+" ");
flago++;
}
if(output.charAt(i)=='u' && flagu==0) {
System.out.print(input.charAt(i)+" ");
flagu++;
}
}
}

public static void main(String args[]) {
String sentence;
int v=0,c=0,ws=0;
Scanner sc= new Scanner(System.in);
sentence = sc.nextLine();
sc.close();
sentence.toLowerCase();
String res="";
for(int i=0;i<sentence.length();i++) {
if(sentence.charAt(i)=='a'||sentence.charAt(i)=='e'||sentence.charAt(i)=='i'||sentence.charAt(i)=='o'||sentence.charAt(i)=='u') {
if(res.indexOf(sentence.charAt(i))<0) {
res+=sentence.charAt(i);
v++;
}//System.out.println(res.indexOf(sentence.charAt(i)));
}
else if(sentence.charAt(i)==' ')
ws++;
else c++;
}
System.out.println(res);
System.out.println("no of vowels: "+v+"\n"+"no of consonants: "+c+"\n"+"no of
white spaces: "+ws);
}

You can use this Method to Find Count of Distinct vowels.
public static int count_Vowels(String str) {
char[] c = str.toLowerCase().toCharArray();
int Counter=0;
String NewString="";
for(int i=0;i<c.length;i++){
String tempString="";
tempString+=c[i];
if(!NewString.contains(tempString) && (c[i]=='a'||c[i]=='e'||c[i]=='i'||c[i]=='o'||c[i]=='u')){
Counter++;
NewString+=c[i];
}
}
return Counter;
}

Here is a solve for this problem without using objects. It's a crude but great solve for beginners who encounter this problem with limited js experience.
How to count unique vowels is a string;
function hasUniqueFourVowels(str){
let va = 0
let ve = 0
let vi = 0
let vo = 0
let vu = 0
let sum = 0
for(let i = 0; i < str.length; i++){
let char = str[i];
if(char === "i"){
vi = 1
}
if(char === "e"){
ve = 1
}
if(char === "a"){
va = 1
}
if(char === "o"){
vo = 1
}
if(char === "u"){
vu = 1
}
sum = va + vi + vo + ve + vu
if (sum >= 4){
return true
}
}
return false
}

#Test
public void numfindVoweles(){
String s="aeiouaedtesssiou";
char a[]=s.toCharArray();
HashMap<Character,Integer> hp= new HashMap<Character, Integer>();
for(char ch:a){
if(hp.containsKey(ch) && (ch=='a' || ch=='e' || ch=='i' || ch=='o' || ch=='u')){
hp.put(ch,hp.get(ch)+1);
}
else if(ch=='a' || ch=='e' || ch=='i' || ch=='o' || ch=='u'){
hp.put(ch,1);
}
}
System.out.println(hp);
}

Use existing Method instead of new Parameter

I'm trying to finish this project and I can't figure how to use my existing method in my other method. I want to get rid of VOWELS, which is defined as a field the class, and I just want to use the method isVowel which returns a boolean after you type in a Char.
This is what I have:
public class StringAndIO {
private static Scanner v;
static final String VOWELS = "AaEeIiOoUuÄäÖöÜü";
public static boolean isVowel(char c) {
if (c == 'a' || c == 'A' || c == 'e' || c == 'E' || c == 'i' || c == 'I' || c == 'o' || c == 'O' || c == 'u'
|| c == 'U' || c == 'ä' || c == 'Ä' || c == 'ö' || c == 'Ö' || c == 'ü' || c == 'Ü') {
return true;
} else {
return false;
}
}
public static String toPigLatin(String text) {
String ret = "";
String vowelbuf = "";
for (int i = 0; i < text.length(); ++i) {
char x = text.charAt(i);
if (VOWELS.indexOf(x) != -1) {
vowelbuf += x;
} else {
if (vowelbuf.length() > 0) {
ret += vowelbuf + "b" + vowelbuf + x;
vowelbuf = "";
} else {
ret += x;
}
}
}
if (vowelbuf.length() > 0) {
ret += vowelbuf + "b" + vowelbuf;
}
return ret;
}
/**
* only there for testing purpose
*/
public static void main(String[] args) {
v = new Scanner(System.in);
System.out.println("Enter a Char!");
char c = v.next().charAt(0);
System.out.println(isVowel(c));
String s = "Meine Mutter ißt gerne Fisch";
System.out.println(s);
System.out.println(toPigLatin(s));
System.out.println();
}
}

THis is how to use your isVowel(x) method inside the other method
public static String toPigLatin(String text) {
String ret = "";
String vowelbuf = "";
for (int i = 0; i < text.length(); ++i) {
char x = text.charAt(i);
if (isVowel(x)) {
vowelbuf += x;
} else {
if (vowelbuf.length() > 0) {
ret += vowelbuf + "b" + vowelbuf + x;
vowelbuf = "";
} else {
ret += x;
}
}
}
if (vowelbuf.length() > 0) {
ret += vowelbuf + "b" + vowelbuf;
}
return ret;
}

Translating a regex into a more basic code

How can I write the below code without using regex?
public static boolean validateCode(String code){
boolean hasAtLeastOneNumber = Pattern.compile("[0-9].*[0-9]")
.matcher(code).find();
boolean hasAtLeastTwoLetters = Pattern.compile("[a-zA-Z].*[a-zA-Z]")
.matcher(code).find();
boolean hasAtLeastOneHyphen = Pattern.compile("-")
.matcher(code).find();
}

How about
public static boolean validateCode2(String code) {
int numbers = 0, letters = 0, hyphens = 0;
for (char c : code.toCharArray()) {
if (Character.isDigit(c)) numbers++;
if (Character.isAlphabetic(c)) letters++;
if (c=='-') hyphens++;
}
return numbers>=2 && letters>=2 && hyphens>=1;
}

For hasAtLeastOneNumber:
for (char c : code.toCharArray()) {
if (Character.isDigit(c)) {
return true;
}
return false;
For hasAtLeastTwoLetters:
int numFound = 0;
for (char c : code.toCharArray()) {
if (Character.isLetter(c)) {
numFound++;
if (numFound >= 2) {
return true;
}
}
}
return false;
For hasAtLeastOneHyphen:
for (char c : code.toCharArray()) {
if (c == '-') {
return true;
}
}
return false;
If you don't want to use toCharArray, you could use:
for (int i=0; i<code.length(); i++) {
char c = code.charAt(i);
// do the rest of the test here
}
That's basically equivalent to using toCharArray except that it's slightly more confusing: someone who looks at the code would need to take a second or two to figure it out. With toCharArray it's obvious what you're doing.

You can loop through the string and test it for ranges of characters. See an example on IDEONE, or ask me if you need an explanation.
import java.util.*;
import java.lang.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
System.out.println(validarCodigo("No-numbers"));
System.out.println(validarCodigo("1-A"));
System.out.println(validarCodigo("This 1 Matches -- :-)"));
}
public static boolean validarCodigo(String codigo) {
int i;
char[] chars = codigo.toCharArray();
char current;
boolean tieneAlmenosUnNumero = false;
boolean tieneAlmenosDosLetras = false;
boolean tieneAlmenosUnGuion = false;
// Check for at least one number
for (i=0; i<chars.length; i++) {
current = chars[i];
if (current >= '0' && current <= '9') {
tieneAlmenosUnNumero = true;
break;
}
}
// Check for at least two letters
int found = 0;
for (i=0; i<chars.length; i++) {
current = chars[i];
boolean lower = current >= 'a' && current <= 'z';
boolean upper = current >= 'A' && current <= 'Z';
if (lower || upper) found++;
if (found == 2){
tieneAlmenosDosLetras = true;
break;
}
}
// Check for at least one hyphen
for (i=0; i<chars.length; i++) {
current = chars[i];
if (current == '-') {
tieneAlmenosUnGuion = true;
break;
}
}
return tieneAlmenosUnNumero && tieneAlmenosDosLetras && tieneAlmenosUnGuion;
}
}

Categories