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

java

python

python3

mysql

mssql

oraclesql

csharp

javascript

typescript

bash

swift

kotlin

dart

golang

ruby

scala

rust

racket

erlang

elixir

pythondata

postgresql

Algorithms

You are given a 0-indexed string <code>num</code> of length <code>n</code> consisting of digits.

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

&nbsp;
Example 1:

<pre>
Input: num = &quot;1210&quot;
Output: true
Explanation:
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>

Example 2:

<pre>
Input: num = &quot;030&quot;
Output: false
Explanation:
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>

&nbsp;
Constraints:

<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>