<p>Given a string <code>s</code>, you&nbsp;can transform every letter individually to be lowercase or uppercase to create another string.</p>

<p>Return <em>a list of all possible strings we could create</em>. Return the output in <strong>any order</strong>.</p>

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

<pre>
<strong>Input:</strong> s = &quot;a1b2&quot;
<strong>Output:</strong> [&quot;a1b2&quot;,&quot;a1B2&quot;,&quot;A1b2&quot;,&quot;A1B2&quot;]
</pre>

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

<pre>
<strong>Input:</strong> s = &quot;3z4&quot;
<strong>Output:</strong> [&quot;3z4&quot;,&quot;3Z4&quot;]
</pre>

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

<ul>
	<li><code>1 &lt;= s.length &lt;= 12</code></li>
	<li><code>s</code> consists of lowercase English letters, uppercase English letters, and digits.</li>
</ul>

class Solution {
public:
    void solve(int i,int &n,string &s,vector<string> &ans){
        if(i>=n){
            ans.push_back(s);
            return;
        }
        if(s[i]>='0'&&s[i]<='9'){
            solve(i+1,n,s,ans);
        }else{
            if(s[i]>='A'&&s[i]<='Z'){
                solve(i+1,n,s,ans);
                s[i] = s[i]-'A'+'a';
                solve(i+1,n,s,ans);
                s[i] = s[i]-'a'+'A';
            }else{
                solve(i+1,n,s,ans);
                s[i] = s[i]+'A'-'a';
                solve(i+1,n,s,ans);
                s[i] = s[i]+'a'-'A';
            }
        }
    }
    vector<string> letterCasePermutation(string s) {
        vector<string> ans;
        int n = s.length();
        solve(0,n,s,ans);
        return ans;
    }
};
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