<p>Given a string containing digits from <code>2-9</code> inclusive, return all possible letter combinations that the number could represent. Return the answer in <strong>any order</strong>.</p>

<p>A mapping of digits to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.</p>
<img alt="" src="https://assets.leetcode.com/uploads/2022/03/15/1200px-telephone-keypad2svg.png" style="width: 300px; height: 243px;" />
<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> digits = &quot;23&quot;
<strong>Output:</strong> [&quot;ad&quot;,&quot;ae&quot;,&quot;af&quot;,&quot;bd&quot;,&quot;be&quot;,&quot;bf&quot;,&quot;cd&quot;,&quot;ce&quot;,&quot;cf&quot;]
</pre>

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

<pre>
<strong>Input:</strong> digits = &quot;&quot;
<strong>Output:</strong> []
</pre>

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

<pre>
<strong>Input:</strong> digits = &quot;2&quot;
<strong>Output:</strong> [&quot;a&quot;,&quot;b&quot;,&quot;c&quot;]
</pre>

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

<ul>
	<li><code>0 &lt;= digits.length &lt;= 4</code></li>
	<li><code>digits[i]</code> is a digit in the range <code>[&#39;2&#39;, &#39;9&#39;]</code>.</li>
</ul>

class Solution {
public:
    unordered_map<char, unordered_set<char>> dic = {
        {'2', {'a', 'b', 'c'}},
        {'3', {'d', 'e', 'f'}},
        {'4', {'g', 'h', 'i'}},
        {'5', {'j', 'k', 'l'}},
        {'6', {'m', 'n', 'o'}},
        {'7', {'p', 'q', 'r', 's'}},
        {'8', {'t', 'u', 'v'}},
        {'9', {'w', 'x', 'y', 'z'}}
    };
    vector<string> letterCombinations(string digits) {
        if (digits == "") {
            return {};
        }
        vector<string> res;
        dfs(digits, 0, "", res);
        return res;
    }
    void dfs(const string& digits, int idx, string str, vector<string> &res) {
        if (idx >= digits.size()) {
            res.push_back(str);
            return ;
        }
        for (auto ch : dic[digits[idx]]) {
            str.push_back(ch);
            dfs(digits, idx+1, str, res);
            str.pop_back();
        }
    }
};
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