What exactly does ++array[s.charAt(i) - 'A'] do?

Question

for (int i = 0; i < s.length(); ++i) 
    {
        if (s.charAt(i) >= 'A' && s.charAt(i) <= 'Z') 
        {
            ++array[s.charAt(i) - 'A'];
        }
    }

I understand the For loop. the s.length() is 26, int[26] to be exact. so this loop will occur 26 times, 0-25. If the Char at i, 0-25 is between or are A-Z it will then proceed to ++array[s.charAt(i) - 'A']; From what i see it adds array once per loop, or adds the value of array once per loop, for the String at char i so the first one would be 0 second would be 2, because arrays start at 0. so adding an array at location of i -'A' is where i get confused.

Solution 1

The statement ++array[s.charAt(i) - 'A']; is incrementing the value in the array indexed by s.charAt(i) - 'A'.

What this loop does is that it counts up the number of occurrences of each letter in s.

The reason for - 'A', is that it "shifts" the ascii/unicode value so that A - Z have values 0 - 25. And are thus more suitable as an array index.

Solution 2

array seems to be a "counter per capital letter". By subtracting character 'A' from an arbitrary character in a string, you get the letter's index in the array:

'A' - 'A' == 0
'B' - 'A' == 1
'C' - 'A' == 2

To understand this, you should understand, that Java treats char the same as (unsigned) short. Hence, you can make calculations with char

Solution 3

What is count[str.charAt(i)]++ actually storing? [duplicate] for those guys who have the same related question here is the answer

First

static final int chars=256; static char count[]=new char[chars];

when we change this code into image it become like this

0 1 2 3 4 5 6 7 8 9 .. .. .. .. .. .. .. 97 .. 255

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

Second

count[str.charAt(i)]++;

Let say you have str = “abc” Then str.charAt(0) will be “a” Then count[‘a’] means count[97] why? Because java automatically converted ‘a’ into ASCII code number 97; Then count[97] value is 0 when you increment it like this count[97]++ then it will become 1

0 1 2 3 4 5 6 7 8 9 .. .. .. .. .. .. .. 97 .. 255

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0

if you increment it again like count[97]++ then it will become 2.

0 1 2 3 4 5 6 7 8 9 .. .. .. .. .. .. .. 97 .. 255

0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0

That is the whole the secret behind. I do not describe the rest because I think your question is answered. If you want to test what I said is correct check it using this simple code

public static void main(String[] argv) {

 int[] count = new int[255]; 
    count['a']++;
    System.out.println(count['a']);
    int counter=0;
    for(int i=0;i<count.length;i++) {
        System.out.println(counter++ + " " + count[i]);
    }

}

Solution 4

count chars

(count chars is not a historical figure)

Solution 5

++array[s.charAt(i) - 'A'];

count frequency of each character in given string s and store the frequency of each character at a particular index in the array.

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