Hey coderz, today we will be solving **Digit Frequency in C HackerRank Solution**.

Contents

**Objective**

Given a string, s, consisting of alphabets and digits, find the frequency of each digit in the given string.

**Input Format**

The first line contains a string, num which is the given number.

**Constraints**

1 <= len(num) <= 1000

All the elements of num are made of english alphabets and digits.

**Output Format**

Print ten space-separated integers in a single line denoting the frequency of each digit from 0 to 9.

**Sample Input 0**

`a11472o5t6`

**Sample Output 0**

`0 2 1 0 1 1 1 1 0 0 `

**Explanation 0**

In the given string:

- 1 occurs two times.
- 2,4,5,6 and 7 occur one time each.
- The remaining digits 0, 3, 8 and 9 don’t occur at all.

**Sample Input 1**

```
lw4n88j12n1
```

**Sample Output 1**

```
0 2 1 0 1 0 0 0 2 0
```

**Sample Input 2**

```
1v88886l256338ar0ekk
```

**Sample Output 2**

`1 1 1 2 0 1 2 0 5 0 `

**Solution – Digit Frequency in C HackerRank Solution**

#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> int main() { char s; int i,a[] ={0,0,0,0,0,0,0,0,0,0}; while(scanf("%c", &s) == 1) if(s >= '0' && s <= '9') a[s-'0']+=1; for(i=0;i<10;i++) printf("%d ",a[i]); return 0; }

**EXPLANATION**

- Declare an frequency array, arr of size 10 and initialize it with zeros, which will be used to count the frequencies of each of the digits occurring.

- Given a string, s, iterate through each of the characters in the string. Check if the current character is a number or not.
- If the current character is a number, increase the frequency of that position in the arr array by 1.
- Once done with the iteration over the string, s, in a new line print all the 10 frequencies starting from 0 to 9, separated by spaces.

**Disclaimer:** The above Problem (**Digit Frequency in C**) is generated by **Hacker Rank** but the Solution is provided by **CodingBroz**.

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