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CRUXEval / sample_83 / 'qq0tt'

Problem

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code

Code

def f(text):
    l = text.rpartition('0')
    if l[2] == '':
        return '-1:-1'
    return f'{len(l[0])}:{l[2].find("0") + 1}'

input

'qq0tt'

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Source and history

upstream_id / Original problem ID: sample_83
task_id: 24dae924-b06c-5a2c-8ba0-7c5ecec00326
task_key: default--test--sample~5f83
task_revision_id: 2
config / split: default / test

Official source

initial import