java - Check whether the given number is fasinating or not

Question

I am doing question on geeksforgeeks and i have come up with a question.

Given a number N. Your task is to check whether it is fascinating or not. Fascinating Number: When a number(should contain 3 digits or more) is multiplied by 2 and 3 ,and when both these products are concatenated with the original number, then it results in all digits from 1 to 9 present exactly once.

Example 1:

Input: N = 192 Output: Fascinating Explanation: After multiplication with 2 and 3, and concatenating with original number, number will become 192384576 which contains all digits from 1 to 9.

Example 2:

Input: N = 853 Output: Not Fascinating Explanation: It's not a fascinating number.

Your Task:
You don't need to read input or print anything. Your task is to complete the function fascinating() which takes the integer n parameters and returns boolean denoting the answer.

Expected Time Complexity: O(1) Expected Auxiliary Space: O(1)

Constraints: 100 <= N <= 107

i have to complete the following fascinating function my code is returning error.

class Solution {
boolean fascinating(String q) {
    // code here
    int sum = 0;
    for(int i=0; i<q.length(); i++){
        sum = sum + Integer.parseInt(Character.toString(q.charAt(i)));
    }
    if(sum == 45){
        return true;
    }
    return false;
}
}

When i saw the solution code i am not able to understand how it is doing the things.

class Solution {
boolean fascinating(String q) {
    int A[] = {0, 0, 0, 0, 0, 0,
               0, 0, 0, 0}; // to store count of every digit from '0' to '9'
    int i, flag = 0;
    char ch;
    for (i = 0; i < q.length(); i++) {
        ch = q.charAt(i);
        A[ch - 48]++;
    }

    for (i = 1; i < 10; i++) {
        // checking if every digit from '1' to '9' are present exactly once
        // or not
        if (A[i] != 1) {
            flag = 1; // flag is set to 1 if frequency is not 1
            break;
        }
    }

    if (flag == 1)
        return false;
    else
        return true;
}
}

Can anyone please explain this to me it will be very helpful.

question from:https://stackoverflow.com/questions/65949337/check-whether-the-given-number-is-fasinating-or-not

Answer 1

class Solution {

fascinating returns a boolean, true/false to the question, "is it fascinating?"

boolean fascinating(String q) {

int A[] = {0, 0, 0, 0, 0, 0,
           0, 0, 0, 0}; // to store count of every digit from '0' to '9'

Here you just initialize A as an array of zeroes, effectively (but not technically) 'empty'.

int i, flag = 0; 
    char ch;

Here you initialize i and flag at 0 and declare ch.

for (i = 0; i < q.length(); i++) {
        ch = q.charAt(i);
        A[ch - 48]++;
    }

Here you have a for loop that goes through each character in q.

For example, let q = 192384576. This is not N, but rather N concatenated with (Nx2) concatenated with (Nx3).

So the for loop first sets ch = q.charAt(0). This is the first character, 1. Then it increments (adds one to) a certain index in A.

This index, much credit and thanks to Andreas for explaining this, is the difference of the Numeric ASCII value for ch and 48, which is the Numeric ASCII value for 0.

In Numeric ASCII:

0 = 48
1 = 49
2 = 50 
3 = 51

and the rest. 

Therefore, ch-48 will give the index in A that should represent ch. For example, when ch is 1, ASCII value of ch is 49 and 49-48 = 1. Therefore, it goes to A[1] and increments it.

Remember, arrays go [0,1,2,3, etc.] So it goes to [0, 0, 0, etc] and adds one to the second 0 -- A[1], thus making A = [0,1,0,etc].

It continues through each digit in q, and thus fills the array A.

for (i = 1; i < 10; i++) {
        // checking if every digit from '1' to '9' are present exactly once
        // or not
        if (A[i] != 1) {
            flag = 1; // flag is set to 1 if frequency is not 1
            break;
        }
    }

Here, it iterates through A, set to length 10 because we are checking 0-9, and checks at each index if the value in that index is equal to 1. Hence, if (A[i]!=1){}

The array should look like [1,1,1,1,etc]. If any of these indices is found not to be 1, it sets flag = 1 and immediately will break or exit the for loop.

if (flag == 1)
        return false;
    else
        return true;
}
}

This final if-else statement checks whether flag is equal to 1. Remember, we only set it equal to 1 if the array is not uniformly filled with 1's, meaning the String q does not contain exactly one of each digit 0-9.

If flag is 1, return false, meaning it is not fascinating. Otherwise, it is fascinating. Thus, when fascinating, return true.