将数组中出现的字符打印到二维数组

Question

I have some characters in char array from az.

1. How can we validate that characters in input array is between az?
2. Print occurrence of every character in a 2D array and also print all the characters between az which are not present in the input array with occurrence count as 0. (Col1 - character, Col2 - occurrence count)

For example:

Input - {'c','b','e','x','h'};

Output -

Col1 Col2

a    0

b    1

c    1

d    0

and so on till z...

I have tried to solve the 1st part of the question as below:

public static boolean charValidation(char[] arr) {

       char[] charArr=new char[122];

        for(char c : arr)
        {
            int ascii = (int)c;
            charArr[ascii]=c;
        }

        for(int i=0;i<charArr.length;i++) {
            if(i<97 || i>122) {
                if((int)charArr[i]!=0) {
                    return false;
                }
            }
        }
        return true;        
    }

It seems to be a hacky solution. Any optimised solution for 1st and 2nd?

Answer 1

Here is a straightforward implementation which populates a hashmap with all lowercase letters of the alphabet, with their counts initialized to zero.

public static boolean charValidation(char[] input) {
    boolean valid = true;
    String letters = "abcdefghijklmnopqrstuvwxyz";
    Map<Character, Integer> counts = new HashMap<>();
    for (char letter : letters.toCharArray()) {
        counts.put(letter, 0);
    }
    for (char letter : input) {
        if (letter < 'a' || letter > 'z') {
            valid = false;
        }
        else {
            int count = counts.get(letter);
            counts.put(letter, count+1);
        }
    }

    counts.entrySet().forEach(entry->{
        System.out.println(entry.getKey() + " " + entry.getValue());  
    });

    return valid;
}

public static void main(String args[]) {
    char[] input = {'c','b','e','x','h'};
    System.out.println("All inputs were valid: " + charValidation(input));
}

This prints:

a 0
b 1
c 1
d 0
e 1
f 0
g 0
h 1
i 0
j 0
k 0
l 0
m 0
n 0
o 0
p 0
q 0
r 0
s 0
t 0
u 0
v 0
w 0
x 1
y 0
z 0
All inputs were valid: true

Answer 2

You can do the first part with just one for loop.

for (int i = 0; i < arr.length; i++)
{
    int ascii = (int) arr[i];

    // if anyone of the ascii codes is outside this range...
    if (ascii < 97 || ascii > 122)
    { 
        // ... then the array has something other than a-z
        // ... so the array is not valid
         return false;
    }
}

// if we made it through the entire for loop,
// this means we never came across an invalid char,
// so return true 

return true;

Answer 3

you can do the required logic in your first loop itself as follows:

        for(char c : arr)
        {
           int ascii = (int)c;
           if(ascii < 97 || ascii > 122)
           {
               return false; 
           }
           return true;
        }

Answer 4

For your 1st questions, how large is your array? If size is not an issue, I would use regexp to achieve it.

Replace anything that is not in az and if the remaining string lengths zero, means all characters are from az.

private static boolean validateChar(char[] arr) {
    if (arr == null || arr.length == 0) {
        return false;
    }
    return new String(arr).replaceAll("[!a-z]", "").length() == 0;
}

For your second question, I think HashMap will do a better job than 2d array, unless you have specific requirement that really needs a 2d array.