# Write a program to check if number has equal digit count and digit value

You are given a 0-indexed string num of length n consisting of digits.

Return true if for every index i in the range 0 <= i < n, the digit i occurs num[i] times in num, otherwise return false.

## Example #1

```Input: num = "1210"
Output: true
Explanation:
num[0] = '1'. The digit 0 occurs once in num.
num[1] = '2'. The digit 1 occurs twice in num.
num[2] = '1'. The digit 2 occurs once in num.
num[3] = '0'. The digit 3 occurs zero times in num.
The condition holds true for every index in "1210", so return true.
```

## Example #2

```Input: num = "030"
Output: false
Explanation:
num[0] = '0'. The digit 0 should occur zero times, but actually occurs twice in num.
num[1] = '3'. The digit 1 should occur three times, but actually occurs zero times in num.
num[2] = '0'. The digit 2 occurs zero times in num.
The indices 0 and 1 both violate the condition, so return false.
```

## The solution:

```    const digitCount = (str) => {
const getFrequency = (string) => {
const freq = {};
for (let i = 0; i < string.length;i++) {
const character = string.charAt(i);
freq[character] ? freq[character]++ : freq[character] = 1;
}
return freq;
}

const freqObj = getFrequency(str);
for (let i = 0; i < str.length;i++) {
const frequency = freqObj[i]? freqObj[i].toString() : '0';
if(str[i] !== frequency) {
return false;
}
}
return true;
}

console.log(digitCount('1210')); // true
```