0005MediumArrayHash TableStringSorting

Bin Label Families

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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 warehouse stamps each storage bin with a short code made of lowercase letters. Two codes belong to the same family when one can be produced by shuffling the letters of the other, which happens exactly when both codes use every letter the same number of times. A code with a letter repeated more often than in another code is therefore in a different family.

Given the array labels, split every code into its family. Each code lands in exactly one family, and repeated codes stay together in the same family, once per occurrence.

Return the families as a list of lists. The families may come back in whatever sequence you like, and the codes inside a family may be arranged however you like.

Examples

Example 1

Input: labels = ["bat", "tab", "cage", "abt", "gaec", "dog"]
Output: [["abt", "bat", "tab"], ["cage", "gaec"], ["dog"]]

`bat`, `tab` and `abt` all use one `a`, one `b` and one `t`, so they share a family; `cage` and `gaec` share another; `dog` has no partner and forms a family alone.

Example 2

Input: labels = ["listen", "silent", "enlist", "tinsel"]
Output: [["enlist", "listen", "silent", "tinsel"]]

Every code draws on the same six letters once each, so a single family holds all four.

Example 3

Input: labels = ["aab", "abb", "aba"]
Output: [["aab", "aba"], ["abb"]]

`aab` and `aba` both carry two `a` and one `b`, while `abb` carries one `a` and two `b`, so counts, not just the set of letters, decide the split.

Constraints

1 <= labels.length <= 10^4
0 <= labels[i].length <= 100
labels[i] consists of lowercase English letters.

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 group_by_letters(labels: list[str]) -> list[list[str]]:

Java

public List<List<String>> groupByLetters(String[] labels)

The intended approach

Time O(n * k log k), where n is the number of codes and k the longest code lengthSpace O(n * k)

Two codes are shuffles of each other precisely when their letters, laid out in sorted order, are identical, so that sorted spelling is a family signature. Bucket every code under its signature in a dictionary and hand back the buckets. Note that a signature has to capture how many times each letter appears, not merely which letters appear, and it cannot be collapsed into a single number: any additive or multiplicative combination of letter values will collide on different multisets.

Pythonentry point: group_by_letters

def group_by_letters(labels):
    families = {}
    for label in labels:
        signature = ''.join(sorted(label))
        if signature not in families:
            families[signature] = []
        families[signature].append(label)
    return list(families.values())

The obvious version

Pairwise multiset comparison with Counter, no signature key at all.

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_groups(labels):
    from collections import Counter
    used = [False] * len(labels)
    out = []
    for i in range(len(labels)):
        if used[i]:
            continue
        group = [labels[i]]
        used[i] = True
        for j in range(i + 1, len(labels)):
            if not used[j] and Counter(labels[j]) == Counter(labels[i]):
                group.append(labels[j])
                used[j] = True
        out.append(group)
    return out

How this problem was checked

Reference solution executed against 12 test cases — 12 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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