<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:
    num_to_char = {'1':[],
                  '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']
                  }
    result = None
    def letterCombinations(self, digits: str) -> List[str]:
        if digits == "":
            return self.result

        self.result = []
        self.backtrack(digits, "")
        return self.result
    
    def backtrack(self, digits, combination):
        if len(digits) == 0:
            self.result.append(combination)
            return
        chars = self.num_to_char[digits[0]]
        for c in chars:
            self.backtrack(digits[1:], combination + c)

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