<p>A <strong>valid IP address</strong> consists of exactly four integers separated by single dots. Each integer is between <code>0</code> and <code>255</code> (<strong>inclusive</strong>) and cannot have leading zeros.</p>

<ul>
	<li>For example, <code>&quot;0.1.2.201&quot;</code> and <code>&quot;192.168.1.1&quot;</code> are <strong>valid</strong> IP addresses, but <code>&quot;0.011.255.245&quot;</code>, <code>&quot;192.168.1.312&quot;</code> and <code>&quot;192.168@1.1&quot;</code> are <strong>invalid</strong> IP addresses.</li>
</ul>

<p>Given a string <code>s</code> containing only digits, return <em>all possible valid IP addresses that can be formed by inserting dots into </em><code>s</code>. You are <strong>not</strong> allowed to reorder or remove any digits in <code>s</code>. You may return the valid IP addresses in <strong>any</strong> order.</p>

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

<pre>
<strong>Input:</strong> s = &quot;25525511135&quot;
<strong>Output:</strong> [&quot;255.255.11.135&quot;,&quot;255.255.111.35&quot;]
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;0000&quot;
<strong>Output:</strong> [&quot;0.0.0.0&quot;]
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;101023&quot;
<strong>Output:</strong> [&quot;1.0.10.23&quot;,&quot;1.0.102.3&quot;,&quot;10.1.0.23&quot;,&quot;10.10.2.3&quot;,&quot;101.0.2.3&quot;]
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 20</code></li>
	<li><code>s</code> consists of digits only.</li>
</ul>

class Solution {
public:
    vector<string> ans;
    vector<string> restoreIpAddresses(string s) {
        helper(s,0,0,s.size(),"");
        return ans;
        
    }

    void helper(string &s,int i,int dc,int n,string temp){
        if(dc==0 && (n-i>12 || n-i<4))
        return;
        if(dc==1 && (n-i>9 || n-i<3))
        return;
        if(dc==2 && (n-i>6 || n-i<2))
        return;
        if(dc==3 && (n-i>3 || n-i<1))
        return;
        

        int x,t=0;
        for(x=0;x<3;x++){
            t=10*t+(s[i]-'0');
            if(t<256){
                
                temp.push_back(s[i]);
                i++;
                if(dc<3){
                temp.push_back('.');
                helper(s,i,dc+1,n,temp);
                temp.pop_back();}
                else{
                    if(i==n){
                        ans.push_back(temp);
                        return;
                    }
                }
                

            }
            if(x==0 && s[i-1]=='0')
            return;
            
        }

    }

    
};
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