check-if-number-has-equal-digit-count-and-digit-value.py

# 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.
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#  
# Example 1:
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# 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:
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# 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.
#
#
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# Constraints:
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# 	n == num.length
# 	1 <= n <= 10
# 	num consists of digits.
#
#


class Solution:
    def digitCount(self, num: str) -> bool:
        for n in range(len(num)):
            if int(num[n]) != num.count(str(n)):
                return False
        return True