About implementing character counting (with.Java)

In this article, we looked at implementing character counting (with.Java).

We’ll be solving a coding test problem, reflecting on the problem we solved, and learning about other ways to solve it.

Let’s start with the problem

Problem

Given a string my_string that contains only alphabetical characters.

count of ‘A’ in my_string, count of ‘B’ in my_string,…, count of ‘Z’ in my_string, count of ‘a’ in my_string, count of ‘b’ in my_string,…

Write a solution function that returns an array of integers of length 52 containing the number of occurrences of ‘z’ in my_string in order.

Example input and output

my_string: “Programmers”

result: [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0]

In my_string, there is one ‘P’, one ‘a’, one ‘e’, one ‘g’, two ‘m’, one ‘o’, three ‘r’, and one ‘s’, so.

returns [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 2, 0, 1, 0, 0, 0, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0].

My solution to the problem

class Solution {
    public int[] solution(String my_string) {
        int[] answer = new int[52];
        for (int i = 0; i < my_string.length(); i++) {
            char c = my_string.charAt(i);

            if (c >= 'A' && c <= 'Z') {
                answer[c - 'A']++;
            } else if (c >= 'a' && c <= 'z') {
                answer[26 + c - 'a']++;
            }
        }

        return answer;
    }
}

Solution

This code counts the frequency of occurrence of alphabetical characters within the given string my_string.

The result is returned as an array of int[], where each element of the array represents the number of occurrences of a particular alphabetical character.

The answer array is an integer array of length 52, storing counts for the uppercase and lowercase letters of the alphabet, respectively.

Indexes 0 through 25 store the number of occurrences of the uppercase letters ‘A’ through ‘Z’.

Indexes 26 through 51 store the number of occurrences of lowercase letters ‘A’ through ‘Z’.

Here’s how the code works:

Iterate over the length of my_string, checking for each letter c.

If c is an uppercase letter, subtract ‘A’ to find the position of that uppercase letter, and increment the count at that position. (answer[c - ‘A’]++)

If c is a lowercase letter, subtract ‘a’ and add 26 to find the corresponding lowercase letter and increment the count at that position. (answer[26 + c - ‘a’]++)

When it has finished processing all the characters, it returns an array of answer. This array represents the number of occurrences of each letter of the alphabet in the string.