<p>You are given positive integers <code>low</code>, <code>high</code>, and <code>k</code>.</p>

<p>A number is <strong>beautiful</strong> if it meets both of the following conditions:</p>

<ul>
	<li>The count of even digits in the number is equal to the count of odd digits.</li>
	<li>The number is divisible by <code>k</code>.</li>
</ul>

<p>Return <em>the number of beautiful integers in the range</em> <code>[low, high]</code>.</p>

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

<pre>
<strong>Input:</strong> low = 10, high = 20, k = 3
<strong>Output:</strong> 2
<strong>Explanation:</strong> There are 2 beautiful integers in the given range: [12,18]. 
- 12 is beautiful because it contains 1 odd digit and 1 even digit, and is divisible by k = 3.
- 18 is beautiful because it contains 1 odd digit and 1 even digit, and is divisible by k = 3.
Additionally we can see that:
- 16 is not beautiful because it is not divisible by k = 3.
- 15 is not beautiful because it does not contain equal counts even and odd digits.
It can be shown that there are only 2 beautiful integers in the given range.
</pre>

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

<pre>
<strong>Input:</strong> low = 1, high = 10, k = 1
<strong>Output:</strong> 1
<strong>Explanation:</strong> There is 1 beautiful integer in the given range: [10].
- 10 is beautiful because it contains 1 odd digit and 1 even digit, and is divisible by k = 1.
It can be shown that there is only 1 beautiful integer in the given range.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> low = 5, high = 5, k = 2
<strong>Output:</strong> 0
<strong>Explanation:</strong> There are 0 beautiful integers in the given range.
- 5 is not beautiful because it is not divisible by k = 2 and it does not contain equal even and odd digits.
</pre>

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

<ul>
	<li><code>0 &lt; low &lt;= high &lt;= 10<sup>9</sup></code></li>
	<li><code>0 &lt; k &lt;= 20</code></li>
</ul>

class Solution:
    def numberOfBeautifulIntegers(self, low: int, high: int, k: int) -> int:
        from functools import lru_cache

        def solve(N):
            N = str(N)
            n = len(N)
            
            from sys import setrecursionlimit
            setrecursionlimit(10000)

            @lru_cache(None)
            def dfs(pos, count_even, count_odd, mod, tight, leading_zero):
                if pos == n:
                    if leading_zero == False and count_even == count_odd and mod == 0:
                        return 1
                    else:
                        return 0
                ans = 0
                max_digit = int(N[pos]) if tight else 9
                for d in range(0, max_digit + 1):
                    new_tight = tight and (d == max_digit)
                    new_leading_zero = leading_zero and (d == 0)
                    new_count_even = count_even
                    new_count_odd = count_odd
                    new_mod = mod
                    if new_leading_zero == False:
                        if d % 2 == 0:
                            new_count_even += 1
                        else:
                            new_count_odd += 1
                        new_mod = (mod * 10 + d) % k
                    ans += dfs(pos + 1, new_count_even, new_count_odd, new_mod, new_tight, new_leading_zero)
                return ans

            return dfs(0, 0, 0, 0, True, True)

        # Compute the count up to high and subtract the count up to low - 1
        result = solve(high) - solve(low - 1)
        return result
        
19200