<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:
        
        @cache
        def dfs(start, ptr, free, mod, b):
            nonlocal n
            if ptr == n:
                return 1 if mod == 0 and b == 0 and start else 0
            ans = 0
            if not start:
                ans = dfs(False, ptr + 1, True, mod, b)
            if free:
                for i in range(10):
                    if i == 0 and not start:
                        continue
                    ans += dfs(True, ptr + 1, True, (mod + i * (10 ** (n - 1 - ptr))) % k, b + o[i % 2])
            else:
                digit = int(arr[ptr])
                for i in range(digit):
                    if i == 0 and not start:
                        continue
                    ans += dfs(True, ptr + 1, True, (mod + i * (10 ** (n - 1 - ptr))) % k, b + o[i % 2])
                ans += dfs(True, ptr + 1, False, (mod + digit * (10 ** (n - 1 - ptr))) % k, b + o[digit % 2])
            return ans

        arr = list(str(high))
        n = len(arr)
        o = [1, -1]
        a1 = dfs(False, 0, False, 0, 0)
        print(a1) # 12
        dfs.cache_clear()
        arr = list(str(low - 1))
        n = len(arr)
        a2 = dfs(False, 0, False, 0, 0)
        print(a2)
        return a1 - a2


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