Check if Number Has Equal Digit Count and Digit Value
<p>You are given a <strong>0-indexed</strong> string <code>num</code> of length <code>n</code> consisting of digits.</p>

<p>Return <code>true</code> <em>if for <strong>every</strong> index </em><code>i</code><em> in the range </em><code>0 &lt;= i &lt; n</code><em>, the digit </em><code>i</code><em> occurs </em><code>num[i]</code><em> times in </em><code>num</code><em>, otherwise return </em><code>false</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;1210&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong>
num[0] = &#39;1&#39;. The digit 0 occurs once in num.
num[1] = &#39;2&#39;. The digit 1 occurs twice in num.
num[2] = &#39;1&#39;. The digit 2 occurs once in num.
num[3] = &#39;0&#39;. The digit 3 occurs zero times in num.
The condition holds true for every index in &quot;1210&quot;, so return true.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;030&quot;
<strong>Output:</strong> false
<strong>Explanation:</strong>
num[0] = &#39;0&#39;. The digit 0 should occur zero times, but actually occurs twice in num.
num[1] = &#39;3&#39;. The digit 1 should occur three times, but actually occurs zero times in num.
num[2] = &#39;0&#39;. The digit 2 occurs zero times in num.
The indices 0 and 1 both violate the condition, so return false.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == num.length</code></li>
	<li><code>1 &lt;= n &lt;= 10</code></li>
	<li><code>num</code> consists of digits.</li>
</ul>
Hash Table; String; Counting
class Solution:
    def digitCount(self, num: str) -> bool:
        cnt = Counter(int(x) for x in num)
        return all(cnt[i] == int(x) for i, x in enumerate(num))
