<p>You are given a string of digits <code>num</code>, such as <code>&quot;123456579&quot;</code>. We can split it into a Fibonacci-like sequence <code>[123, 456, 579]</code>.</p>

<p>Formally, a <strong>Fibonacci-like</strong> sequence is a list <code>f</code> of non-negative integers such that:</p>

<ul>
	<li><code>0 &lt;= f[i] &lt; 2<sup>31</sup></code>, (that is, each integer fits in a <strong>32-bit</strong> signed integer type),</li>
	<li><code>f.length &gt;= 3</code>, and</li>
	<li><code>f[i] + f[i + 1] == f[i + 2]</code> for all <code>0 &lt;= i &lt; f.length - 2</code>.</li>
</ul>

<p>Note that when splitting the string into pieces, each piece must not have extra leading zeroes, except if the piece is the number <code>0</code> itself.</p>

<p>Return any Fibonacci-like sequence split from <code>num</code>, or return <code>[]</code> if it cannot be done.</p>

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

<pre>
<strong>Input:</strong> num = &quot;1101111&quot;
<strong>Output:</strong> [11,0,11,11]
<strong>Explanation:</strong> The output [110, 1, 111] would also be accepted.
</pre>

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

<pre>
<strong>Input:</strong> num = &quot;112358130&quot;
<strong>Output:</strong> []
<strong>Explanation:</strong> The task is impossible.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> num = &quot;0123&quot;
<strong>Output:</strong> []
<strong>Explanation:</strong> Leading zeroes are not allowed, so &quot;01&quot;, &quot;2&quot;, &quot;3&quot; is not valid.
</pre>

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

<ul>
	<li><code>1 &lt;= num.length &lt;= 200</code></li>
	<li><code>num</code> contains only digits.</li>
</ul>

class Solution {
public:
    // you need to be extra concious about the large input beacuse it is given
    // in string format . in java we can handle //that but in c++ we need
    // different technique to handle that....
    bool backtrack(string s, vector<int>& ans, int index, long long f1,
                   long long f2, int size) {
        // base case if we reach end and our ans have more than 2 element than
        // its true otherwise false
        if (s.size() == index)
            return size > 2;
        long long int n = 0;
        for (int i = index; i < s.size(); i++) {
            n = n * 10 + (s[i] - '0');
            // to avoid negative overflow
            if (n < 0)
                return false;
            // to avoid positive overflow
            if (n > INT_MAX)
                break;
            // if we had not reach to 3 elements i.e ans have only two elements
            // than we need to check if we can put this //number into our ans or
            // not
            if (size < 2 || f1 + f2 == n) {
                ans.push_back(n);
                // basic backtrack step to check further cases if it is true
                // then we would return true
                if (backtrack(s, ans, i + 1, f2, n, size + 1))
                    return true;
                // if we are not able to find ans then we would backtrack our
                // answer and thus we would remove this last element
                ans.pop_back();
            }
            if (n == 0) // if n is 0 i.e at start only if n is 0 we would return
                        // false because this would voilate question rules.
                return false;
        }
        return false;
    }
    vector<int> splitIntoFibonacci(string S) {
        vector<int> ans;
        // here we need two to pass three different variable except string
        // itself, ans vector, index i.e 1st numer, 2nd //number and the size of
        // vector to avoid overflow condition because input maybe very very
        // large...
        backtrack(S, ans, 0, 0, 0, 0);
        return ans;
    }
};

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