<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>ans;
    void dfs(string s,int i,string temp)
    {
        
        if(i>s.length()-1)
        {
        ans.push_back(temp);
        temp=s;
        return;
        }

        if(temp[i]>='a' and temp[i]<='z'){
        temp[i]=(char)toupper(temp[i]);
        dfs(s,i+1,temp);
        temp[i]=(char)tolower(temp[i]);
        }
        else if(temp[i]>='A' and temp[i]<='Z'){
        temp[i]=(char)tolower(temp[i]);
        dfs(s,i+1,temp);
        temp[i]=(char)toupper(temp[i]);
        }
        dfs(s,i+1,temp);
    }
    vector<string> letterCasePermutation(string s) {
        dfs(s,0,s);
        return ans;
    }
};
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