<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:
    vector<int> ans;
    void dfs(int i,string &num,vector<int> &dp,int n){
        
        //base case
        
      if(i>=n and dp.size()>=3){
          
          ans= dp;
          return;
      }
        
        if(i>=n)return;
        
        //recursive case
        
        for(int j=i;j<i+10;j++){
            
            auto left= num.substr(i,j-i+1);
            
            long long leftNum= stoll(left);
            
            if(leftNum > INT_MAX)return;
            int size= dp.size();
            
            if(j==i and num[j]=='0'){
                
                 if(dp.size()>=2 and (long long)dp[size-1] + (long long)dp[size-2] == leftNum){
                
                dp.push_back(leftNum);
                dfs(j+1,num,dp,n);
                dp.pop_back();
            }
            
            if(dp.size()<2){
                
                dp.push_back(leftNum);
                dfs(j+1,num,dp,n);
                dp.pop_back();
                }
                
                break;
            }
            if(dp.size()<2){
                
                dp.push_back(leftNum);
                dfs(j+1,num,dp,n);
                dp.pop_back();
            }
            
            if(dp.size()>=2 and (long long)dp[size-1] + (long long)dp[size-2] == leftNum){
                
                dp.push_back(leftNum);
                dfs(j+1,num,dp,n);
                dp.pop_back();
            }
            
            
        }
            
           
    }
    vector<int> splitIntoFibonacci(string num) {
        // if(num[0]=='0')return {};
        
        int n= num.length();
        
        vector<int> dp;
        dfs(0,num,dp,n);
        
        
        return ans;
    }
};
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