# 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