Occurrences in char Array

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I have this problem, Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 100 at Exercise_12_2.main(Exercise_12_2.java:28) When i am trying to just simply count the occurrences of the letters in a char array. I just cant seem to wrap my head around how to work it out. I have been at it for several hours. Please help get me on the right track.

input: a a a b b c !
Expected output:

Counts:
a    3
b    2
c    1

This is my code so far. Please help me.

import java.util.Scanner;

public class Exercise_12_2
{
    public static void main(String[] args)
    {
        Scanner input = new Scanner(System.in); // Setup scanner

        char[] charArray = new char[100];
        int[] counts = new int[26];
        char tempinput = '?';

        System.out.print("Enter letters (or ! to quit): ");

        while (tempinput != '!')
        {
            tempinput = (input.next()).charAt(0);
            charArray[tempinput]++;

            for (int c = 'a'; c <= 'z'; c++)
            {
                for (int k = 0; k <= charArray.length; k++)
                {
                    if (c == charArray[k])
                    {
                        counts[c] += 1;
                    }
                }
            }
        }
    }
    public static void displayCounts(int[] counts)
    {
        for (int i = 0; i < counts.length; i++)
        {
            if ((i + 1) % 10 == 0)
            {
                System.out.println(counts[i] + " " + (char)(i + 'a'));
            }
            else
            {
                System.out.print(counts[i] + " " + (char)(i + 'a') + " ");
            }
        }
    }
}

java.lang.ArrayIndexOutOfBoundsException: 100 means you are asking for the 101st element of an array with only 100 elements. Remember, java arrays are indexed starting with 0.

From that you might be able to tell why this line is broken:

for (int k = 0; k <= charArray.length; k++)

charArray.length is 100, so you run up until k is 101, meaning that you try charArray[100] which is asking for the 101st element of charArray, but charArray only has 100 elements.

If you switch k <= charArray.length to k < charArray.length you should get farther.

On an unrelated note, counts[c] += 1; won't work. The int val of a char is its ascii value. The letter a for example is 97, so you'll go way off the end of your counts array. This also doesn't account for upper/lower case (which have different numeric values).

Java program to find the occurrence of a character in a string, We are then creating an another array array to hold the characters of the given String, then we are comparing them with the characters in the String and when a​  what you intended in that regard, but with just a little additional effort, you can use a separate index and a separate array to store the first instance of each character (in say a chars[] array) along with the count of each in your count[] array and preserve your inorder count while eliminating duplicate characters. The changes needed are

This problem becomes much easier if you use an appropriate data structure. For example, I would use a Map<Character, Integer> to keep the count and then iterate that to display. Like,

Scanner input = new Scanner(System.in); // Setup scanner
System.out.print("Enter letters (or ! to quit): ");
Map<Character, Integer> map = new HashMap<>();
char ch;
while ((ch = Character.toLowerCase(input.next().charAt(0))) != '!') {
    map.put(ch, map.getOrDefault(ch, 0) + 1);
}
System.out.println("Counts:");
for (int c = 'a'; c <= 'z'; c++) {
    if (map.containsKey((char) c)) {
        System.out.printf("%c\t%d%n", c, map.get((char) c));
    }
}

Which I tested with your example input, and I get as requested:

Enter letters (or ! to quit): a a a b b c !
Counts:
a   3
b   2
c   1

Java program to count the occurrences of each character , Write a Java program which prints number of occurrences of each characters and also it should not print The idea is to create a count array of size 256. Occurrences of each char individually With forEach and a MutableMap : val charsMap = mutableMapOf<Char, Int>() string2.forEach{ charsMap[it] = charsMap.getOrDefault(it, 0) + 1 } print(charsMap)

I think after creating the character array of [a, a, b c, d, !, ..], following code will be enough to calculate count for each alphabet [a-z].

for (int k = 0; k < charArray.length; k++)
    {
        char c = charArray[k];
        if (c >= 'a' && c <= 'z' )
        {
            counts[c - 'a'] += 1;
        }
    }

Program to count occurrence of a given character in a string , Program to count occurrence of a given character in a string. Given a C++ program to count occurrences of a given. // character int count(string s, char c). Count number of occurrences (or frequency) in a sorted array Given a sorted array arr[] and a number x, write a function that counts the occurrences of x in arr[]. Expected time complexity is O(Logn)

Count Occurrences of a Char in a String, Learn how to count characters with the core Java library and with libraries and frameworks such as Spring and Guava. There are many ways to count the number of occurrences of a char in a String in Java. In this quick article, we'll focus on a few examples of how to count characters, first, with the core Java library and then with other libraries and frameworks such as Spring and Guava.

Count occurrences of character in String java program code with , Then , we will traverse char array and increment its value by 1 for each Character present in the input string. public class EachCharacterCountInString { public  Formula to Count the Number of Occurrences of a Single Character in a Range =SUM(LEN( range )-LEN(SUBSTITUTE( range ,"a",""))) Where range is the cell range in question, and "a" is replaced by the character you want to count.

Occurrences in char Array, Occurrences in char Array. occurrence of each character in string java how to count the number of occurrences of a character in a string in c count occurrences of  class JavaExample { static void countEachChar(String str) { //ASCII values ranges upto 256 int counter[] = new int[256]; //String length int len = str.length(); /* This array holds the occurrence of each char, For example * ASCII value of A is 65 so if A is found twice then * counter[65] would have the value 2, here 65 is the ASCII value * of A */ for (int i = 0; i < len; i++) counter[str.charAt(i)]++; // We are creating another array with the size of String char array[] = new char[str

Comments
  • THANK YOU THANK YOU. This helped a ton. I feel blind. Makes much more sense now. I really appreciate it
  • @IsaacPeterson, please read the edit starting with "On an unrelated note". Also I would appreciate it if you would accept the answer :)
  • I appreciate the answer, It works. I just haven't learned about Map and HashMap stuff yet. I will refer back to this when I do.
  • I will definitely give this a try. Thanks!