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(Java) Counting letters in a sentence?

The following is my code:
char[] array = new char[26] ;
int index = 0 ;
int letter = 0 ;
int countA = 0 ;
String sentence = "Once upon a time..." ;
if(sentence.contains("."))
{
String sentenceNoSpace = sentence.replace(" ", "").toLowerCase() ;
String sentenceFinal = sentenceNoSpace.substring(0, sentenceNoSpace.indexOf(".")) ;
char[] count = new char[sentenceFinal.length()] ;
for (char c = 'a'; c <= 'z'; c++)
{
array[index++] = c ;
for(int i = 0; i < sentenceFinal.length(); i++)
{
if(sentenceFinal.charAt(i) == c)
count[letter++] = c ;
//if(sentenceFinal.charAt(i) == 'a')
//countA++ ;
}
}
String result = new String(count) ; // Convert to a string.
System.out.println("\n" + result) ;
System.out.println("\nTotal number of letters is " + result.length()) ;
System.out.println(countA) ;
}
else
{
System.out.println("You forgot a period. Try again.") ;
}
I am having trouble counting how many a's, b's, c's, etc. are in a given sentence. There is one way I can do it and it is this part
//if(sentenceFinal.charAt(i) == 'a')
//countA++ ;
which I can just create all the way to z. Is there a more efficient way?
Note: No using Hashmap or any other advance techniques.

There is no need of eliminating spaces. This is just additional work you're doing.
int countOfLetters = 0 ;
String sentence = "Once upon a time..." ;
sentence = sentence.toLowerCase();
int[] countOfAlphabets = new int[26];
for (int i = 0; i < sentence.length(); i++) {
if (sentence.charAt(i) >= 'a' && sentence.charAt(i) <= 'z') {
countOfAlphabets[sentence.charAt(i) - 97]++;
countOfLetters++;
}
}
So, countOfLetters will give you the total count of letters.
If you want individual count, suppose for example, you want count of 'c',
You can get it by accessing countOfAlphabets array like countOfAlphabets['c' - 97] (97 being the ASCII value of 'a')

Use an int array letterCounts that will store the counts for each letter. Assuming the case of the letters can be ignored, the length of the letterCounts array will be 26.
Iterate over the string's characters and update the corresponding integer in the array. Use its ASCII value to find the corresponding index, as follows.
letterCounts[c - 97]++
97 is the ASCII value of 'a', whose count needs to be stored at index 0.
In this way, subtracting 97 from the character's ASCII value will give the corresponding index for that character.
Note: This is assuming that you want to store the counts for lowercase letters.

Pretty fiddly without using maps, but this will count all characters in a string.
You might want to modify to exclude things like spaces etc.
public class Main {
public static void main(String[] args) {
String sentence = "Once upon a time...";
// Create an array of size 256 ASCII_SIZE
int count[] = new int[256];
int length = sentence.length();
// Initialize count array index
for (int i = 0; i < length; i++)
count[sentence.charAt(i)]++;
// Create an array of given String size
char chars[] = new char[sentence.length()];
for (int i = 0; i < length; i++) {
chars[i] = sentence.charAt(i);
int find = 0;
for (int j = 0; j <= i; j++) {
// If any matches found
if (sentence.charAt(i) == chars[j])
find++;
}
if (find == 1) {
System.out.println("Occurrence of " + sentence.charAt(i) + " is:" + count[sentence.charAt(i)]);
}
}
}
}
Which outputs:
Occurrence of O is:1
Occurrence of n is:2
Occurrence of c is:1
Occurrence of e is:2
Occurrence of is:3
Occurrence of u is:1
Occurrence of p is:1
Occurrence of o is:1
Occurrence of a is:1
Occurrence of t is:1
Occurrence of i is:1
Occurrence of m is:1
Occurrence of . is:3

Check Below code You can have a 26 length array and index will increment according to the presence of the alphabet.
public void getResult(){
int [] charCount = new int [26];
int countA = 0 ;
String sentence = "Once upon a time..." ;
if(sentence.contains("."))
{
String sentenceNoSpace = sentence.replace(" ", "").toLowerCase() ;
String sentenceFinal = sentenceNoSpace.substring(0, sentenceNoSpace.indexOf(".")) ;
char[] sentenceCharArray = sentenceFinal.toCharArray();
//char a = 97;
for (int i = 0; i <sentenceCharArray.length ; i++) {
int index = sentenceCharArray[i] - 97 ;
if(index >= 0 && index <= 26) {
charCount[index] += 1;
}
}
System.out.print("Result : ");
for (int i = 0; i < charCount.length ; i++) {
System.out.print(charCount [i]+" , ");
}
System.out.println("\nTotal number of letters is " + sentenceCharArray.length) ;
}
else
{
System.out.println("You forgot a period. Try again.") ;
}
}

Since there are 26 letters in the US alphabet, you can use an int[] with a size of 26
int[] letterCount = new int[26];
to hold the count of each letter where index 0 represents 'a', 1 represents 'b', etc...
As you traverse through the sentence, check if the character you're on is a letter, Character.isLetter(), then increment the element in the array that represents the letter.
letterCount[letter - 'a']++;
We subtract 'a' from the letter to give us the correct index.
Code Sample
package stackoverflow;
public class Question {
public static void main(String[] args) {
String sentence = "The quick brown fox jumps over the lazy dog.";
int[] letterCount = new int[26];
if (sentence.contains(".")) {
// toLowerCase() the sentence since we count upper and lowercase as the same
for (char letter : sentence.toLowerCase().toCharArray()) {
if (Character.isLetter(letter)) {
letterCount[letter - 'a']++;
}
}
// Display the count of each letter that was found
int sumOfLetters = 0;
for (int i = 0; i < letterCount.length; i++) {
int count = letterCount[i];
if (count > 0) {
System.out.println((char)(i + 'a') + " occurs " + count + " times");
sumOfLetters += count;
}
}
System.out.println("Total number of letters is " + sumOfLetters);
} else {
System.out.println("You forgot a period. Try again.");
}
}
}
Result
a occurs 1 times
b occurs 1 times
c occurs 1 times
d occurs 1 times
e occurs 3 times
f occurs 1 times
g occurs 1 times
h occurs 2 times
i occurs 1 times
j occurs 1 times
k occurs 1 times
l occurs 1 times
m occurs 1 times
n occurs 1 times
o occurs 4 times
p occurs 1 times
q occurs 1 times
r occurs 2 times
s occurs 1 times
t occurs 2 times
u occurs 2 times
v occurs 1 times
w occurs 1 times
x occurs 1 times
y occurs 1 times
z occurs 1 times
Total number of letters is 35
Rebuttal Question
What is wrong with using Java 8 and using the chars() of a String? With it, you can accomplish the same thing with less code. For the total number of letters, we just use String.replaceAll() and remove all non-letters from the String with the pattern [^A-Za-z]and use the length() of the result.
package stackoverflow;
import java.util.function.Function;
import java.util.stream.Collectors;
public class Question {
public static void main(String[] args) {
String sentence = "The quick brown fox jumps over the lazy dog.";
System.out.println(sentence.toLowerCase().chars()
// Change the IntStream to a stream of Characters
.mapToObj(c -> (char)c)
// Filter out non lower case letters
.filter(c -> 'a' <= c && c <= 'z')
// Collect up the letters and count them
.collect(Collectors.groupingBy(Function.identity(), Collectors.counting())));
System.out.println("Total letter count is " + sentence.replaceAll("[^A-Za-z]", "").length());
}
}
Result
{a=1, b=1, c=1, d=1, e=3, f=1, g=1, h=2, i=1, j=1, k=1, l=1, m=1, n=1, o=4, p=1, q=1, r=2, s=1, t=2, u=2, v=1, w=1, x=1, y=1, z=1}
Total letter count is 35

You can solve it with Regex If Regex wont be considered as High-tech 🙂
Idea is simple: Remove all letters and subtract output from original string length to get counter
String sentence = "Once upon a time...";
String noLetterString = sentence.replaceAll("[a-zA-Z]", "");
int counterLetter = sentence.length() - noLetterString.length();
System.out.println("counter:" + counterLetter);
By old school programming 🙂
Idea here is in reverse, appending only letters
String sentence = "Once upon a time...";
String lowerCase = sentence.toLowerCase(); // to avoid comparison to UpperCase letters
StringBuilder counterStr = new StringBuilder();
for (char l : lowerCase.toCharArray()) {
if (l >= 'a' && l <= 'z') {
counterStr.append(l);
}
}
System.out.println("counterStr:" + counterStr);
System.out.println("counter:" + counterStr.length());

Here is the Update Code :
int[] array = new int[26] ;
String sentence = "Once upon a time..." ;
if(sentence.contains("."))
{
String sentenceNoSpace = sentence.replace(" ", "").toLowerCase() ;
String sentenceFinal = sentenceNoSpace.substring(0, sentenceNoSpace.indexOf(".")) ;
for (char c : sentenceFinal.toCharArray())
{
System.out.println(c+" "+(c-97));
array[c-97] += 1;
}
// System.out.println("\n" + Arrays.toString(array)) ;
for(int i=0; i< array.length;i++) {
if(array[i] != 0) {
char c = (char)(i+97);
System.out.println(c+" occured "+ array[i]+" times");
}
}
}
else
{
System.out.println("You forgot a period. Try again.") ;
}

Java/Android: how to find the first characters of a string if it equals other string

I have a String Array #1 countries[] and I have another String Array #2 splitr_str[] that have the same values
but it split in half.
What I want is to to check which value of the #2 Array match the first half value of the #1 String Array which has the full word.
Here's an example:
String[] countries = new String[]{"USA", "Spain" ,"Germany"};
String[] split_str = new String[]{"Sp", "ain", "A" ,"US" ,"Ger", "ma","ny"};
Now i want to detect only the first half of the countries values that match the ssplit_str value
for USA i just want to pick "US" which is in
split_str[3]
and if i found it ? i pick "A" which is
split_str[2]
And so on.
Any ideas on how to do that?

Try using for loops. As I just learned from this post, you an use a label and break to easily break out of both loops once the match has been found.
outerloop:
for( String country : countries){
for( String segment : split_str) {
if( country.startsWith( segment ) ) {
// do something here when the segment has been detected
break outerloop;
}
}
}

I know its a bit inefficient. But its the best solution I could find
for (int i = 0; i < countries.length; i++)
{
for(int j = 0; j < split_str.length; j++)
{
if(countries[i].substring(0, (countries[i].length() / 2) + 1).contains(split_str[j]))
{
for (int k = 0; k < split_str.length; k++)
{
if (countries[i].substring((countries[i].length() / 2) + 1, countries[i].length()).contains(split_str[k]))
{
System.out.println(i);
}
}
}
}
}
Hope it helps!

Unsure how to implement for loop

Hello I am having trouble implementing this function
Function:
Decompress the String s. Character in the string is preceded by a number. The number tells you how many times to repeat the letter. return a new string.
"3d1v0m" becomes "dddv"
I realize my code is incorrect thus far. I am unsure on how to fix it.
My code thus far is :
int start = 0;
for(int j = 0; j < s.length(); j++){
if (s.isDigit(charAt(s.indexOf(j)) == true){
Integer.parseInt(s.substring(0, s.index(j))

Assuming the input is in correct format, the following can be a simple code using for loop. Of course this is not a stylish code and you may write more concise and functional style code using Commons Lang or Guava.
StringBuilder builder = new StringBuilder();
for (int i = 0; i < s.length(); i += 2) {
final int n = Character.getNumericValue(s.charAt(i));
for (int j = 0; j < n; j++) {
builder.append(s.charAt(i + 1));
}
}
System.out.println(builder.toString());

Here is a solution you may like to use that uses Regex:
String query = "3d1v0m";
StringBuilder result = new StringBuilder();
String[] digitsA = query.split("\\D+");
String[] letterA = query.split("[0-9]+");
for (int arrIndex = 0; arrIndex < digitsA.length; arrIndex++)
{
for (int count = 0; count < Integer.parseInt(digitsA[arrIndex]); count++)
{
result.append(letterA[arrIndex + 1]);
}
}
System.out.println(result);
Output
dddv
This solution is scalable to support more than 1 digit numbers and more than 1 letter patterns.
i.e.
Input
3vs1a10m
Output
vsvsvsammmmmmmmmm

Though Nami's answer is terse and good. I'm still adding my solution for variety, built as a static method, which does not use a nested For loop, instead, it uses a While loop. And, it requires that the input string has even number of characters and every odd positioned character in the compressed string is a number.
public static String decompress_string(String compressed_string)
{
String decompressed_string = "";
for(int i=0; i<compressed_string.length(); i = i+2) //Skip by 2 characters in the compressed string
{
if(compressed_string.substring(i, i+1).matches("\\d")) //Check for a number at odd positions
{
int reps = Integer.parseInt(compressed_string.substring(i, i+1)); //Take the first number
String character = compressed_string.substring(i+1, i+2); //Take the next character in sequence
int count = 1;
while(count<=reps)//check if at least one repetition is required
{
decompressed_string = decompressed_string + character; //append the character to end of string
count++;
};
}
else
{
//In case the first character of the code pair is not a number
//Or when the string has uneven number of characters
return("Incorrect compressed string!!");
}
}
return decompressed_string;
}

Loop through char array and count chars

I'm trying to optimize my Java code so I try things. In my search for a short way I produced the code here below. This throws an Exception. Can you tell me why? Can't I loop through a string by a char Array?
public class Input {
public static void main(String[] args) {
String check = "Dit moet toch gewoon te doen zijn !!";
check = check.toLowerCase();
int[] counter = {0, 0, 0, 0, 0};
char[] vowel = {'a', 'e', 'i', 'o', 'u'};
int total = 0;
for (int i = 0; i < check.length(); i++)
if (check.charAt(i) == vowel[i])
counter[i]++;
for (int t : counter)
total += t;
System.out.println("Aantal klinkers: \t" + total);
}
}

Your code reads like this:
For each character in "check"
if character at index in "check" is character at index in "vowel"
That's probably not what you're looking for. The exception you're getting is because there are only 5 characters in "vowel" but alot in "check" (i'm not counting)
Now, I'm assuming what you're wanting to do is actually count the number of each vowel in "check"
You should actually use a nested for loop in this case.
for (int i = 0; i < check.length(); i++) {
for (int v = 0; v < vowel.length; v++) {
if (check.charAt(i) == vowel[v]) {
counter[v]++;
break;
}
}
}

for (int i = 0; i < check.length(); i++)
if (check.charAt(i) == vowel[i])
counter[i]++;
This loop goes from 0 to check.length(); but your array vowel[] has 5 element. So it generate array out of bound exception.

One more for the clever regex department so you don't have to make your string lowercase and lose the cases for vowels:
if (check.matches("(?i)[^a-z][a-z].*"))
System.out.println("Vowels found: " + check.replaceAll("[^a-z]+", ""));

You must have two loops
One for going through each char in string and a for loop going through each vowel array like so
for (int i = 0; i < check.length(); i++)
for (int p = 0; p < vowel.length; p++)
if (check.charAt(i) == vowel[p])
counter[p]++;
Enjoy.

You could use a regex for this operation:
Pattern vowels = Pattern.compile("(?i)[aeiou]");
String check = "Dit moet toch gewoon te doen zijn !!";
Matcher matcher = vowels.matcher(check);
while(matcher.find()) {
System.out.println("found vowel: " + matcher.group());
}
Explanation of regex:
(?i) make pattern case insensitive, [aeiou] characters to match.

Letter frequency array, converting int to char

My code below prints out a letter frequency table (number of letter occurances in a string) that is inputted from a scanner class. I Have everything working correctly except the last bit. When executed, the table displays letters a-z and the number of occurrences except for the last method LargestLength() method. Since my array 'score' uses int[26], instead of using a-z it uses ints 0-25. (0=A, 1=B, etc.) As of now my LargestLength method only displays the number (instead of the letter) which comes up the most but does not count the number of occurrences.
For example, If my string were to be "Hello", l shows up most frequently so i would like it to display, "Most Frequent: l 2" but with my code it displays "Most Frequent: 11" (11 = L). How would I go about fixing this?
Profile Class.
public class LetterProfile {
int score[] = new int [26];
public void countChars (String s) {
s = s.toLowerCase();
char a = 'a';
for (int i = 0; i < s.length(); i++) {
int next = (int)s.charAt(i) - (int) a;
if ( next< 26 && next >= 0)
score[next]++;
}
}
public int largestLength() {
int largest = 0;
int largestindex = 0;
for(int a = 0; a<26; a++) {
if(score[a] > largest) {
largest = score[a];
largestindex = a ;
}
}
return (char) largestindex;
}
public void printResults() {
largestLength();
for (int i = 0; i < score.length; i++) {
System.out.println( (char)(i+97) + ": " + score[i]);
}
System.out.println(("Most Frequent") + (": ") + largestLength());
}
}
A bit of a confusing explanation but any help would be appreciated.

If my string were to be "Hello", l shows up most frequently so i would like it to display, "Most Frequent: l 2" but with my code it displays "Most Frequent: 11" (11 = L). How would I go about fixing this?
Simple: String.valueOf((char)(number + 'A'));