0031MediumHash TableStringBacktracking

Keypad Code Spellings

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Trains the technique from

LeetCode 17Letter Combinations of a Phone Number

This is an original problem, written from a brief that listed the technique, the difficulty, the topics, the function shape and the input bounds — none of that problem's wording, examples, hints or editorials. The link is there so you can map your practice onto the standard set.

Same function shape, different story and different numbers.

A parcel locker is opened with a word, but its keypad is the old telephone kind: every key from 2 to 9 carries a small group of lowercase letters, and the keys 0 and 1 carry none.

2 -> a b c      6 -> m n o
3 -> d e f      7 -> p q r s
4 -> g h i      8 -> t u v
5 -> j k l      9 -> w x y z

A courier watched a customer press the keys but could not see which letter on each key was intended. Given the pressed sequence as a string code, list every word that could have been meant, taking exactly one letter from the group of each key, in the order the keys were pressed.

You may list the words in whatever order you like. When code is empty nothing was pressed, so the answer is an empty list.

Examples

Example 1

Input: code = "79"
Output: ["pw", "px", "py", "pz", "qw", "qx", "qy", "qz", "rw", "rx", "ry", "rz", "sw", "sx", "sy", "sz"]

Key 7 offers four letters and key 9 offers four, giving 4 * 4 = 16 candidate words.

Example 2

Input: code = "94"
Output: ["wg", "wh", "wi", "xg", "xh", "xi", "yg", "yh", "yi", "zg", "zh", "zi"]

Four letters on key 9 pair with three on key 4, so twelve words fit the sequence.

Example 3

Input: code = "4"
Output: ["g", "h", "i"]

One key was pressed, so each of its three letters is a candidate on its own.

Constraints

0 <= code.length <= 4
Every character of code is a digit from '2' through '9'.

The groups you return, and the values inside each group, may be in any order.

The signature

The editor is preloaded with this. It matches the parent problem's shape, so a solution that works here transfers to a judge unchanged.

Python

def keypad_spellings(code: str) -> list[str]:

Java

public List<String> keypadSpellings(String code)

The intended approach

Time O(n * 4^n)Space O(n) for the buffer and recursion, plus the returned list

Recurse over the pressed keys, holding one partial word in a shared buffer. At depth i try each letter of key i, push it, recurse, then pop it before trying the next letter; when the depth passes the last key, record the buffer as a finished word. The empty sequence short-circuits before any recursion so nothing is recorded.

Pythonentry point: keypad_spellings

def keypad_spellings(code):
    groups = {
        "2": "abc", "3": "def", "4": "ghi", "5": "jkl",
        "6": "mno", "7": "pqrs", "8": "tuv", "9": "wxyz",
    }
    if not code:
        return []
    words = []
    buf = []

    def walk(i):
        if i == len(code):
            words.append("".join(buf))
            return
        for letter in groups[code[i]]:
            buf.append(letter)
            walk(i + 1)
            buf.pop()

    walk(0)
    return words

The obvious version

Hand every key's letter group to a cartesian-product helper and join each tuple it yields, no recursion of our own.

This is not filler. It was written independently of the fast solution and the two were run against thousands of random inputs to confirm they agree — which is how the answer above is known to be right. Read it when the intended approach feels like a rabbit pulled out of a hat.

Python

def bf_keypad_spellings(code):
    from itertools import product
    groups = {
        "2": "abc", "3": "def", "4": "ghi", "5": "jkl",
        "6": "mno", "7": "pqrs", "8": "tuv", "9": "wxyz",
    }
    if len(code) == 0:
        return []
    pools = [groups[d] for d in code]
    return ["".join(combo) for combo in product(*pools)]

How this problem was checked

Reference solution executed against 8 test cases — 8 passed
3 worked examples confirmed by running the code
0 eight-word sequences shared with the parent problem's text
Agrees with an independent brute-force implementation on random inputs
Test set proven to catch deliberately broken solutions
Reviewed by a separate quality pass
Function arity matches the parent problem's

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