<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;
    void solve(string &s , int dotcnt  , int prevdot , string str, int n ){
        if(dotcnt == 4 && prevdot == -1){
            reverse(str.begin() , str.end());
            ans.push_back(str);
        }

        if(dotcnt > 3 || prevdot < 0)return;
    
        int i;
        int cnt = 0;
        int val = 0;
        int power = 1;
        for(i = prevdot;i>=0 && cnt < 3 ; i--){
            val += (s[i] - '0')*power;
            power *= 10;
            str.push_back(s[i]);
            if(cnt == 0 && val >= 0){
                

                if(dotcnt < 3)
                str.push_back('.');

                solve(s , dotcnt + 1 ,i-1 , str,n );

                if(dotcnt < 3)
                str.pop_back();

            }else if(cnt == 1 && val >=10){

                if(dotcnt < 3)
                str.push_back('.');

                solve(s , dotcnt + 1 ,i-1 , str,n );

                if(dotcnt < 3)
                str.pop_back();

            }else if(cnt == 2 && val >= 100 && val <= 255){

                if(dotcnt < 3)
                str.push_back('.');

                solve(s , dotcnt + 1 ,i-1 , str,n );

                if(dotcnt < 3)
                str.pop_back();
            } 
            cnt++; 
        }
    }
    vector<string> restoreIpAddresses(string s) {
        //we can do recursion for solving this problem 
        //only three dots are allowed

        int n = s.size();
        if(n < 4)return ans;//if string size is less than 4 no ip exist.
        string str;
        solve(s , 0 ,n-1 , str,n );
        return ans; 
    }
};
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