<p>Given an integer <code>n</code>, count <em>the total number of digit </em><code>1</code><em> appearing in all non-negative integers less than or equal to</em> <code>n</code>.</p>

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

<pre>
<strong>Input:</strong> n = 13
<strong>Output:</strong> 6
</pre>

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

<pre>
<strong>Input:</strong> n = 0
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>0 &lt;= n &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def countDigitOne(self, n: int) -> int:
        def recurse(s):
            greatest10 = (len(s)-1)*10**(len(s)-2)
            if(len(s) == 1):
                return 1 if s != "0" else 0
            if(s[0] == "1"):
                return greatest10 + int(s[1:]) + recurse(str(int(s[1:]))) + 1
            return greatest10*int(s[0]) + 10**(len(s)-1) + recurse(str(int(s[1:])))
        
        return recurse(str(n))

16400