<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>

class Solution:
  def countSteppingNumbers(self, low: str, high: str) -> int:
    def is_stepping(num: str) -> bool:
      for i in range(1, len(num)):
        if abs(ord(num[i]) - ord(num[i - 1])) != 1:
          return False
      return True
    
    mod = 1000000007
    def helper(num: str) -> int:
      n = len(num)
      @cache
      def dfs(i: int, previous: int, has_constraints: bool, started: bool) -> int:
        if i == n:
          return 1
        
        digit = ord(num[i]) - ord('0')
        end_digit = digit + 1 if has_constraints else 10
        ret = 0
        for d in range(end_digit):
          if started and abs(previous - d) != 1:
            continue
          
          ret = (ret + dfs(i + 1, d, has_constraints and d == digit, started or d != 0))
        return ret
      
      return dfs(0, -1, True, False)
    
    return ((mod + helper(high) - helper(low)) % mod + (1 if is_stepping(low) else 0)) % mod
      
20800