I got stuck in this syntax of c++. "Check if the frequency of all the digits in a number is same"

Question

I'm practicing a coding problem on "Check if the frequency of all the digits in a number is same"

#include<bits/stdc++.h> 
using namespace std; 

    // returns true if the number 
    // passed as the argument 
    // is a balanced number. 

bool isNumBalanced(int N) 
{ 

    string st = to_string(N); 
    bool isBalanced = true; 

    // frequency array to store 
    // the frequencies of all 
    // the digits of the number 
    int freq[10] = {0}; 
    int i = 0; 
    int n = st.size(); 

    for (i = 0; i < n; i++) 

        // store the frequency of 
        // the current digit 
        freq[st[i] - '0']++; 

    for (i = 0; i < 9; i++) 
    { 

        // if freq[i] is not 
        // equal to freq[i + 1] at 
        // any index 'i' then set 
        // isBalanced to false 
        if (freq[i] != freq[i + 1]) 
            isBalanced = false; 
    } 

    // return true if 
    // the string is balanced 
    if (isBalanced) 
        return true; 
    else
        return false; 
} 

// Driver code 
int main() 
{ 
    int N = 1234567890; 
    bool flag = isNumBalanced(N); 

    if (flag) 
        cout << "YES"; 
    else
        cout << "NO"; 
} 

but I can't understand this code:

// store the frequency of
// the current digit
freq[st[i] - '0']++;

How this part actually working and storing frequency? And instead of this line, what else I can write?

Answer 1

st is a string and thus, a sequence of chars. st[i] is the ith char in this string.

Chars are actually positive integers between 0 and 256, so you can use them with mathematical operations, such as -. These integers are assigned to characters according to the ASCII alphabet. For example: The char 0 is assigned to 48 and the char 7 to 55 (Note: in the following, I use x to denote the character).

Their order makes it possible that mathematical operations are sensible as follows: The char 7 and the char 0 are exactly 7 numbers apart, so 0 + 7 = 48 + 7 = 55 = 7. So: 7 - 0 = 7.

So, you get the position in the freq array according to the number, i.e., the position 0 for 0 or position 7 for 7. The ++ operator increments that value in-place.

Answer 2

This line is several things condensed into one expression

freq[st[i] - '0']++;

The individual part are rather simple and in total it also isn't too difficult:

st[i] - '0' - character digits do not map 1 to 1 to integers. There is an offset. The integer value of '1' is 1 + '0', '2' is 2 + '0'. Hence to get the integer from the digit you need to subtract '0'.

freq[ ... ] - accesses the element of the array. Element at index i stores frequency of digit i.

()++ - increments that frequency by one.

Answer 3

Subtracting the '0' character from the single string character results in the actual number you're looking for. This gives you the number whose frequency you are tracking in your code. This works because of the way characters are stored as ASCII values. Check out the table below. Say that the integer value N that is passed in is 1221. The first value observed in this example is '1' which corresponds to an ASCII value of 49. The ASCII value of '0' is 48. Subtracting the two: 49 - 48 = 1. This allows you to access each integer value individually as part of the array that was the result of the transformation of an 'int' value into a string.

ASCII Table

Answer 4

The code of

for (i = 0; i < n; i++) 

    // store the frequency of 
    // the current digit 
    freq[st[i] - '0']++; 

traverses the string and for each item, it subtracts '0', which has a value of 48, because character code 48 represents 0, character code 49 represents 1 and so on.

This code however is superfluos, wastes memory in storing a string and wastes time converting a number to a string. This is better:

bool isNumBalanced(int N)
{
    //We create an array of 10 for each digit
    int digits[10];
    //Initialize the difits
    for (int i = 0; i < 10; i++) digits[i] = 0;
    //If the input is 0, then we have a trivial case
    if (N == 0) return true;
    //We loop the digits
    do {
        //N % 10 is the last digit
        //We increment the frequency of that digit
        digits[N % 10]++;
    } while ((N /= 10) != 0); //We don't stop until we reach the trivial case, see above
    //Using the transitivity of equality, we compare all values to the first
    //We return false upon the first difference
    for (int j = 1; j < 10; j++)
        if (digits[0] != digits[j]) return false;
    //Otherwise we return true
    return true;
}

Answer 5

For those who don't understand it.

int arr[5]={0} // it stores 0 in all places
for(int i=0;i<5;i++){
  arr[i]++;
} // Now the array is [1 1 1 1 1]

what happened here is first i=0 then arr[0]++ "here arr[0] value was 0, ++, it increment 0 to 1" now arr[0] value is 1. Now ` let

   st="1221";

   for (i = 0; i < 4; i++) {

     freq[st[i] - '0']++;



for i=0, the freq location is : freq[49-48]++ = freq[1]++ means value of freq[1] is 1
for i=1, the freq location is : freq[50-48]++ = freq[2]++ means value of freq[2] is 1
for i=2, the freq location is : freq[50-48]++ = freq[2]++ means value of freq[2] is 2
for i=3, the freq location is : freq[49-48]++ = freq[1]++ means value of freq[1] is 2

ASCII value of '0' is 48

ASCII value of '1' is 49

ASCII value of '2'is 50