<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:
        
        numsPath = []
        def dfs(start):
            if start >= len(num) and len(numsPath) > 2:
                return True
            if start >= len(num) and len(numsPath) <= 2:
                return False
            

            for i in range(start, len(num)):
                if len(numsPath) < 2:
                    possibleNum = num[start : i + 1]
                    # is this not a valid num that does not start with 0
                    if str(int(possibleNum)) != possibleNum:
                        continue
                        
                    numsPath.append(possibleNum)
                    if (dfs(i + 1)):
                        return True

                elif len(numsPath) >= 2:
                    possibleNum = num[start : i + 1]

                    # is this not a valid num that does not start with 0
                    if str(int(possibleNum)) != possibleNum:
                        continue

                    # is this num a sum of the previous two nums
                    n1 = int(numsPath[len(numsPath) - 2])
                    n2 = int(numsPath[len(numsPath) - 1])
                    n3 = n1 + n2

                    if int(possibleNum) != n3:
                        continue
                    
                    numsPath.append(possibleNum)
                    if (dfs(i + 1)):
                        return True

                numsPath.pop()
   
        return dfs(0)

            
16400