<p>Given two positive integers <code>low</code> and <code>high</code> represented as strings, find the count of <strong>stepping numbers</strong> in the inclusive range <code>[low, high]</code>.</p>

<p>A <strong>stepping number</strong> is an integer such that all of its adjacent digits have an absolute difference of <strong>exactly</strong> <code>1</code>.</p>

<p>Return <em>an integer denoting the count of stepping numbers in the inclusive range</em> <code>[low, high]</code><em>. </em></p>

<p>Since the answer may be very large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p><strong>Note:</strong> A stepping number should not have a leading zero.</p>

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

<pre>
<strong>Input:</strong> low = &quot;1&quot;, high = &quot;11&quot;
<strong>Output:</strong> 10
<strong>Explanation: </strong>The stepping numbers in the range [1,11] are 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. There are a total of 10 stepping numbers in the range. Hence, the output is 10.</pre>

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

<pre>
<strong>Input:</strong> low = &quot;90&quot;, high = &quot;101&quot;
<strong>Output:</strong> 2
<strong>Explanation: </strong>The stepping numbers in the range [90,101] are 98 and 101. There are a total of 2 stepping numbers in the range. Hence, the output is 2. </pre>

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

<ul>
	<li><code>1 &lt;= int(low) &lt;= int(high) &lt; 10<sup>100</sup></code></li>
	<li><code>1 &lt;= low.length, high.length &lt;= 100</code></li>
	<li><code>low</code> and <code>high</code> consist of only digits.</li>
	<li><code>low</code> and <code>high</code> don&#39;t have any leading zeros.</li>
</ul>

from functools import cache

class Solution:
    def countSteppingNumbers(self, low: str, high: str) -> int:
        MOD = 10 ** 9 + 7
        def solve(R):
            nr = len(R)
            @cache
            def dfs(n, prev, tight, prezero):
                if n == 0:
                    return 1
                ubound = ord(R[nr - n]) - ord('0') if tight else 9
                res = 0
                if prev == -1 or prezero:
                    for dig in range(ubound + 1):
                        res += dfs(n - 1, dig, tight and (dig == ubound), prezero and (dig == 0))
                else:
                    for dig in range(ubound + 1):
                        if abs(dig - prev) == 1:
                            res += dfs(n - 1, dig, tight and (dig == ubound), prezero and (dig == 0))
                return res % MOD
            return dfs(nr, -1, 1, 1)
        return (solve(high) - solve(str(int(low) - 1))) % MOD
20700