<p>An <strong>additive number</strong> is a string whose digits can form an <strong>additive sequence</strong>.</p>

<p>A valid <strong>additive sequence</strong> should contain <strong>at least</strong> three numbers. Except for the first two numbers, each subsequent number in the sequence must be the sum of the preceding two.</p>

<p>Given a string containing only digits, return <code>true</code> if it is an <strong>additive number</strong> or <code>false</code> otherwise.</p>

<p><strong>Note:</strong> Numbers in the additive sequence <strong>cannot</strong> have leading zeros, so sequence <code>1, 2, 03</code> or <code>1, 02, 3</code> is invalid.</p>

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

<pre>
<strong>Input:</strong> &quot;112358&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> 
The digits can form an additive sequence: 1, 1, 2, 3, 5, 8. 
1 + 1 = 2, 1 + 2 = 3, 2 + 3 = 5, 3 + 5 = 8
</pre>

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

<pre>
<strong>Input:</strong> &quot;199100199&quot;
<strong>Output:</strong> true
<strong>Explanation:</strong> 
The additive sequence is: 1, 99, 100, 199.&nbsp;
1 + 99 = 100, 99 + 100 = 199
</pre>

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

<ul>
	<li><code>1 &lt;= num.length &lt;= 35</code></li>
	<li><code>num</code> consists only of digits.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> How would you handle overflow for very large input integers?</p>

class Solution:
  def isAdditiveNumber(self, num: str) -> bool:
    n = len(num)

    def dfs(firstNum: int, secondNum: int, s: int) -> bool:
      if s == len(num):
        return True

      thirdNum = firstNum + secondNum
      thirdNumStr = str(thirdNum)

      return (num.find(thirdNumStr, s) == s and
              dfs(secondNum, thirdNum, s + len(thirdNumStr)))

    # num[0..i] = firstNum
    for i in range(n // 2):
      if i > 0 and num[0] == '0':
        return False
      firstNum = int(num[:i + 1])
      # num[i + 1..j] = secondNum
      # |thirdNum| >= max(|firstNum|, |secondNum|)
      j = i + 1
      while max(i, j - i) < n - j:
        if j > i + 1 and num[i + 1] == '0':
          break
        secondNum = int(num[i + 1:j + 1])
        if dfs(firstNum, secondNum, j + 1):
          return True
        j += 1

    return False
16500