0959MediumHash TableStringSorting

Rearranging a Label to a Given Letter Order

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Trains the technique from

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 letter order is given as order, a string of distinct lowercase letters, and a label as s.

Rearrange the letters of s so that any two letters both appearing in order come in the same relative order as they do there. Letters of s not appearing in order all come after those that do, arranged alphabetically among themselves.

Return the rearranged label. Every letter of s is kept, repeats included.

Examples

Example 1

Input: order = "fedcba", s = "abcdefghij"
Output: "fedcbaghij"

The order reverses the first six letters, so those come out backwards, and the four letters not named there follow alphabetically.

Example 2

Input: order = "ba", s = "aabbcc"
Output: "bbaacc"

The order asks for b before a, so both b's come first, then both a's, and the unnamed c's follow.

Example 3

Input: order = "abc", s = "zyx"
Output: "xyz"

None of the label's letters are named in the order, so they simply come out alphabetically.

Constraints

1 <= order.length <= 26
1 <= s.length <= 200
order and s consist of lowercase English letters only
The letters of order are all different

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 custom_sort_string(order: str, s: str) -> str:

Java

public String customSortString(String order, String s)

The intended approach

Time O(n + A) for a label of length n over an alphabet of size A.Space O(A) for the tally.

Nothing about the label's original arrangement survives, so only how often each letter appears matters. Tally the letters of the label once, then write the answer out: walk `order` and, for each letter, write it as many times as the tally says, which puts the ordered letters exactly where the order demands and keeps repeats together. Then walk the alphabet and write out whatever is left over, which is every letter not named in `order`, alphabetically by construction.

Pythonentry point: custom_sort_string

def custom_sort_string(order, s):
    tally = [0] * 26
    for ch in s:
        tally[ord(ch) - 97] += 1
    out = []
    named = [False] * 26
    for ch in order:
        at = ord(ch) - 97
        named[at] = True
        out.append(ch * tally[at])
    for at in range(26):
        if named[at]:
            continue
        if tally[at]:
            out.append(chr(97 + at) * tally[at])
    return "".join(out)

The obvious version

Give each letter a rank, its position in the order for the letters named there and a rank past the end of the order for the rest, then sort the label's letters by that rank with the letter itself breaking ties. Sorts the characters rather than writing them out from a tally.

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_custom_sort_string(order, s):
    rank = {}
    for i in range(len(order)):
        rank[order[i]] = i
    letters = list(s)
    letters.sort(key=lambda ch: (rank.get(ch, len(order) + 1), ch))
    return "".join(letters)

How this problem was checked

Reference solution executed against 10 test cases — 10 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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