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CRUXEval / sample_398 / '2': 2, '0': 1, '1': 2

Problem

Answer published by the source. Consult the official source to check your work against its answer.

code

Code

def f(counts):
    dict = {}
    for k, v in counts.items():
        count = counts[k]
        if count not in dict:
            dict[count] = []
        dict[count].append(k)
    counts.update(dict)
    return counts

input

{'2': 2, '0': 1, '1': 2}

Discussion

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Source and history

upstream_id / Original problem ID
sample_398
task_id
dd05dc9d-6323-5800-8e30-2dfbd03a3e5a
task_key
default--test--sample~5f398
task_revision_id
1
config / split
default / test

Official source

initial import