<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:
    vector<string> letterCasePermutation(string s) {
        vector<string> ans;
        string curS = "";
        int n = s.size();
        auto dfs = [&](auto &&dfs, int cur){
            if(cur == n){
                ans.push_back(curS);
                return;
            }
            if(isalpha(s[cur])){
                curS+=tolower(s[cur]);
                dfs(dfs, cur+1);
                curS.pop_back();
                curS += toupper(s[cur]);
                dfs(dfs, cur+1);
                curS.pop_back();
            }
            else{
                curS+=s[cur];
                dfs(dfs, cur+1);
                curS.pop_back();
            }
            return;
        };
        dfs(dfs, 0);
        return ans;
    }
};
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