<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:
    digit_to_alphabets_map = {
        '2': list("abc"),
        '3': list("def"),
        '4': list("ghi"),
        '5': list("jkl"),
        '6': list("mno"),
        '7': list("pqrs"),
        '8': list("tuv"),
        '9': list("wxyz"),
    }

    def letterCombinations(self, digits: str) -> List[str]:
        if not digits:
            return ""

        res = []
        ans = ""

        self.backtrack(digits, ans, res, next_idx=0)
        return res

    def backtrack(self, digits, ans, res, next_idx) -> None:
        if len(ans) == len(digits):
            res.append(ans)
        
        for idx, digit in enumerate(digits):
            if idx < next_idx:
                continue
            
            for alphabet in self.digit_to_alphabets_map[digit]:
                ans += alphabet
                self.backtrack(digits, ans, res, next_idx=idx + 1)
                ans = ans[:-1]

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