<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 {
    using LL = long long;
public:
    bool isAdditiveNumber(string num) {
        int n = num.size();
        auto dfs= [&](auto&& dfs, int i, LL n1, LL n2, int cnt) {
            if(i == n) {
                return cnt >= 3;
            }
            for(int j=i; j<(num[i] == '0' ? i+1 : min(n,i+18)); j++) {
                LL x = stoll(num.substr(i, j-i+1));
                if(n1 == -1 || n2 == -1 || x == n1 + n2) {
                    if(dfs(dfs, j+1, n2, x, cnt+1)) return true;
                }
            }
            return false;
        };
        return dfs(dfs, 0, -1, -1, 0);
    }
};
9200