<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 {
 public:
  bool isAdditiveNumber(string num) {
    const int n = num.length();

    // num[0..i] = firstNum
    for (int i = 0; i < n / 2; ++i) {
      if (i > 0 && num[0] == '0')
        return false;
      const long firstNum = stol(num.substr(0, i + 1));
      // num[i + 1..j] = secondNum
      // |thirdNum| >= max(|firstNum|, |secondNum|)
      for (int j = i + 1; max(i, j - i) < n - j; ++j) {
        if (j > i + 1 && num[i + 1] == '0')
          break;
        const long secondNum = stol(num.substr(i + 1, j - i));
        if (dfs(num, firstNum, secondNum, j + 1))
          return true;
      }
    }

    return false;
  }

 private:
  bool dfs(const string& num, long firstNum, long secondNum, long s) {
    if (s == num.length())
      return true;

    const long thirdNum = firstNum + secondNum;
    const string& thirdNumStr = to_string(thirdNum);
    return num.find(thirdNumStr, s) == s &&
           dfs(num, secondNum, thirdNum, s + thirdNumStr.length());
  }
};
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