Run Detail
completedbeam-1.0-501b-a23b30 answersRecorded for these videos
Provider
openrouter
Challenge Version
v1.0
Turns
30
Duration
46.2s
Total Tokens
55,910
Input Tokens
5,034
Output Tokens
50,876
Reasoning Tokens
46,922
Reasoning Effort
default
Cached Input Tokens
2,432
Est. Cost
$0.0516
Started
11/6/2026, 12:08:41 PM
OR Provider
Nebius
Finish Reason
stop
Generation ID
gen-1793966949-Io9YI3POpuSy2cpTnBNJ
Turns (30)
Turn #1Raw score: 1
121 in234 out182 reasoning355 total0.8s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write is_leap(year: int) -> bool to return whether year is a leap year in the Gregorian calendar.
assistant
def is_leap(year: int) -> bool:
return (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
Turn #2Raw score: 1
121 in180 out128 reasoning64 cached301 total0.9s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write is_leap(year: int) -> bool to return whether year is a leap year in the Gregorian calendar.
assistant
def is_leap(year: int) -> bool:
return (year % 4 == 0 and year % 100 != 0) or (year % 400 == 0)
Turn #3Raw score: 1
121 in415 out365 reasoning64 cached536 total1.5s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write is_leap(year: int) -> bool to return whether year is a leap year in the Gregorian calendar.
assistant
def is_leap(year: int) -> bool:
return year % 4 == 0 and (year % 100 != 0 or year % 400 == 0)
Turn #4Raw score: 1
157 in288 out187 reasoning445 total0.8s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write rle(s: str) -> str to perform run-length encoding on s. Replace each run of consecutive identical characters with that character followed by the run length. rle("aaabcc") returns "a3b1c2" and rle("") returns "".
assistant
def rle(s: str) -> str:
if not s:
return ""
result = []
current = s[0]
count = 1
for ch in s[1:]:
if ch == current:
count += 1
else:
result.append(f"{current}{count}")
current = ch
count = 1
result.append(f"{current}{count}")
return "".join(result)
Turn #5Raw score: 1
157 in207 out164 reasoning64 cached364 total0.7s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write rle(s: str) -> str to perform run-length encoding on s. Replace each run of consecutive identical characters with that character followed by the run length. rle("aaabcc") returns "a3b1c2" and rle("") returns "".
assistant
import itertools
def rle(s: str) -> str:
return "".join(f"{char}{len(list(group))}" for char, group in itertools.groupby(s))
Turn #6Raw score: 1
157 in160 out55 reasoning128 cached317 total0.8s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write rle(s: str) -> str to perform run-length encoding on s. Replace each run of consecutive identical characters with that character followed by the run length. rle("aaabcc") returns "a3b1c2" and rle("") returns "".
assistant
def rle(s: str) -> str:
if not s:
return ""
result = []
current = s[0]
count = 1
for ch in s[1:]:
if ch == current:
count += 1
else:
result.append(current)
result.append(str(count))
current = ch
count = 1
result.append(current)
result.append(str(count))
return "".join(result)
Turn #7Raw score: 1
152 in841 out756 reasoning64 cached993 total2.6s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write chunks(items, n: int) -> list[list]. items is an iterable, which may be a generator. Return consecutive lists of n elements, where the last list may be shorter. Raise ValueError when n is less than 1.
assistant
import itertools
def chunks(items, n: int) -> list[list]:
if n < 1:
raise ValueError("n must be at least 1")
iterator = iter(items)
result = []
while True:
chunk = list(itertools.islice(iterator, n))
if not chunk:
break
result.append(chunk)
return result
Turn #8Raw score: 1
152 in358 out270 reasoning64 cached510 total0.9s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write chunks(items, n: int) -> list[list]. items is an iterable, which may be a generator. Return consecutive lists of n elements, where the last list may be shorter. Raise ValueError when n is less than 1.
assistant
def chunks(items, n: int) -> list[list]:
if n < 1:
raise ValueError("n must be at least 1")
result = []
current = []
for item in items:
current.append(item)
if len(current) == n:
result.append(current)
current = []
if current:
result.append(current)
return result
Turn #9Raw score: 1
152 in291 out197 reasoning128 cached443 total1.1s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write chunks(items, n: int) -> list[list]. items is an iterable, which may be a generator. Return consecutive lists of n elements, where the last list may be shorter. Raise ValueError when n is less than 1.
assistant
def chunks(items, n: int) -> list[list]:
if n < 1:
raise ValueError("n must be at least 1")
result: list[list] = []
current: list = []
for item in items:
current.append(item)
if len(current) == n:
result.append(current)
current = []
if current:
result.append(current)
return result
Turn #10Raw score: 1
190 in6,451 out6,254 reasoning64 cached6,641 total18.2s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write round_half_up(value: str, digits: int) -> str. The string value represents a decimal number. Round it to digits decimal places, rounding exact ties away from zero based on the exact value written in value. Return a string with exactly digits decimal places, and omit the decimal point when digits is 0. round_half_up("2.675", 2) returns "2.68".
assistant
from decimal import Decimal, ROUND_HALF_UP, localcontext
def round_half_up(value: str, digits: int) -> str:
d = Decimal(value)
try:
adjusted = d.adjusted()
except Exception:
adjusted = 0
if d == 0:
adjusted = 0
precision = max(28, adjusted + digits + 3)
if precision < 1:
precision = 1
quantum = Decimal("1e{}".format(-digits))
with localcontext() as ctx:
ctx.prec = precision
emax = max(999999, abs(adjusted) + abs(digits) + 10)
ctx.Emax = emax
ctx.Emin = -emax
rounded = d.quantize(quantum, rounding=ROUND_HALF_UP)
return format(rounded, "f")
Turn #11Raw score: 1
190 in6,427 out6,272 reasoning128 cached6,617 total16.8s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write round_half_up(value: str, digits: int) -> str. The string value represents a decimal number. Round it to digits decimal places, rounding exact ties away from zero based on the exact value written in value. Return a string with exactly digits decimal places, and omit the decimal point when digits is 0. round_half_up("2.675", 2) returns "2.68".
assistant
from decimal import Decimal, localcontext, ROUND_HALF_UP
def round_half_up(value: str, digits: int) -> str:
d = Decimal(value)
if d.is_finite():
prec = max(
28,
len(d.as_tuple().digits) + 5,
d.adjusted() + digits + 2,
digits + 1,
)
else:
prec = 28
with localcontext() as ctx:
ctx.prec = prec
ctx.rounding = ROUND_HALF_UP
unit = Decimal(1).scaleb(-digits)
rounded = d.quantize(unit, rounding=ROUND_HALF_UP)
return str(rounded)
Turn #12Raw score: 1
190 in3,218 out3,098 reasoning64 cached3,408 total9.6s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write round_half_up(value: str, digits: int) -> str. The string value represents a decimal number. Round it to digits decimal places, rounding exact ties away from zero based on the exact value written in value. Return a string with exactly digits decimal places, and omit the decimal point when digits is 0. round_half_up("2.675", 2) returns "2.68".
assistant
from decimal import Decimal, ROUND_HALF_UP, localcontext
def round_half_up(value: str, digits: int) -> str:
d = Decimal(value)
quantizer = Decimal(1).scaleb(-digits)
with localcontext() as ctx:
ctx.prec = max(28, len(d.as_tuple().digits) + abs(d.as_tuple().exponent) + digits + 10)
rounded = d.quantize(quantizer, rounding=ROUND_HALF_UP)
return format(rounded, "f")
Turn #13Raw score: 1
185 in3,280 out2,991 reasoning64 cached3,465 total9.8s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write parse_kv(line: str) -> dict[str, str]. The string line contains key=value pairs separated by one or more spaces. A value may be enclosed in double quotes; if so, it may contain spaces and =, and the quotes are not included in the parsed value. A value may be empty, as in c=. If a key repeats, use the last value. Return {} for an empty line.
assistant
def parse_kv(line: str) -> dict[str, str]:
result: dict[str, str] = {}
n = len(line)
i = 0
while i < n and line[i].isspace():
i += 1
while i < n:
start = i
while i < n and not line[i].isspace() and line[i] != "=":
i += 1
key = line[start:i]
if i < n and line[i] == "=":
i += 1
else:
while i < n and not line[i].isspace():
i += 1
while i < n and line[i].isspace():
i += 1
continue
if i < n and line[i] == '"':
i += 1
start = i
while i < n and line[i] != '"':
i += 1
value = line[start:i]
if i < n and line[i] == '"':
i += 1
else:
start = i
while i < n and not line[i].isspace():
i += 1
value = line[start:i]
result[key] = value
while i < n and line[i].isspace():
i += 1
return result
Turn #14Raw score: 1
185 in2,470 out2,193 reasoning128 cached2,655 total7.6s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write parse_kv(line: str) -> dict[str, str]. The string line contains key=value pairs separated by one or more spaces. A value may be enclosed in double quotes; if so, it may contain spaces and =, and the quotes are not included in the parsed value. A value may be empty, as in c=. If a key repeats, use the last value. Return {} for an empty line.
assistant
def parse_kv(line: str) -> dict[str, str]:
result: dict[str, str] = {}
i = 0
n = len(line)
while i < n and line[i].isspace():
i += 1
while i < n:
key_start = i
while i < n and line[i] != "=" and not line[i].isspace():
i += 1
key = line[key_start:i]
while i < n and line[i].isspace():
i += 1
if i >= n or line[i] != "=":
break
i += 1 # skip '='
if i < n and line[i] == '"':
i += 1
value_start = i
while i < n and line[i] != '"':
i += 1
value = line[value_start:i]
if i < n and line[i] == '"':
i += 1
else:
value_start = i
while i < n and not line[i].isspace():
i += 1
value = line[value_start:i]
result[key] = value
while i < n and line[i].isspace():
i += 1
return result
Turn #15Raw score: 1
185 in1,838 out1,574 reasoning64 cached2,023 total5.7s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write parse_kv(line: str) -> dict[str, str]. The string line contains key=value pairs separated by one or more spaces. A value may be enclosed in double quotes; if so, it may contain spaces and =, and the quotes are not included in the parsed value. A value may be empty, as in c=. If a key repeats, use the last value. Return {} for an empty line.
assistant
def parse_kv(line: str) -> dict[str, str]:
result: dict[str, str] = {}
i = 0
n = len(line)
while i < n and line[i].isspace():
i += 1
while i < n:
key_start = i
while i < n and line[i] != '=' and not line[i].isspace():
i += 1
if i >= n or line[i] != '=':
break
key = line[key_start:i]
i += 1 # skip '='
if i < n and line[i] == '"':
i += 1
value_start = i
while i < n and line[i] != '"':
i += 1
value = line[value_start:i]
if i < n and line[i] == '"':
i += 1
else:
value_start = i
while i < n and not line[i].isspace():
i += 1
value = line[value_start:i]
if key:
result[key] = value
while i < n and line[i].isspace():
i += 1
return result
Turn #16Raw score: 1
171 in384 out265 reasoning64 cached555 total1.2s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write merge_intervals(intervals: list[tuple[int, int]]) -> list[tuple[int, int]]. The input intervals may be in any order. Merge intervals that overlap or touch, so (1, 2) and (2, 3) become (1, 3). Return the merged intervals as tuples sorted by start, and leave the input list unchanged.
assistant
def merge_intervals(intervals: list[tuple[int, int]]) -> list[tuple[int, int]]:
if not intervals:
return []
ordered = sorted(intervals, key=lambda interval: interval[0])
merged = [ordered[0]]
for start, end in ordered[1:]:
last_start, last_end = merged[-1]
if start <= last_end:
merged[-1] = (last_start, max(last_end, end))
else:
merged.append((start, end))
return merged
Turn #17Raw score: 1
171 in599 out474 reasoning64 cached770 total2.5s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write merge_intervals(intervals: list[tuple[int, int]]) -> list[tuple[int, int]]. The input intervals may be in any order. Merge intervals that overlap or touch, so (1, 2) and (2, 3) become (1, 3). Return the merged intervals as tuples sorted by start, and leave the input list unchanged.
assistant
def merge_intervals(intervals: list[tuple[int, int]]) -> list[tuple[int, int]]:
if not intervals:
return []
sorted_intervals = sorted(intervals)
merged: list[tuple[int, int]] = [sorted_intervals[0]]
for start, end in sorted_intervals[1:]:
last_start, last_end = merged[-1]
if start <= last_end:
if end > last_end:
merged[-1] = (last_start, end)
else:
merged.append((start, end))
return merged
Turn #18Raw score: 1
171 in478 out364 reasoning128 cached649 total1.6s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write merge_intervals(intervals: list[tuple[int, int]]) -> list[tuple[int, int]]. The input intervals may be in any order. Merge intervals that overlap or touch, so (1, 2) and (2, 3) become (1, 3). Return the merged intervals as tuples sorted by start, and leave the input list unchanged.
assistant
def merge_intervals(intervals: list[tuple[int, int]]) -> list[tuple[int, int]]:
if not intervals:
return []
sorted_intervals = sorted(intervals)
merged = [sorted_intervals[0]]
for start, end in sorted_intervals[1:]:
last_start, last_end = merged[-1]
if start <= last_end:
merged[-1] = (last_start, max(last_end, end))
else:
merged.append((start, end))
return merged
Turn #19Raw score: 1
147 in608 out500 reasoning64 cached755 total2.1s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write add_months(d: datetime.date, months: int) -> datetime.date to add months calendar months to d. months may be negative. If the day does not exist in the target month, use the last day of that month.
assistant
import calendar
import datetime
def add_months(d: datetime.date, months: int) -> datetime.date:
total_months = d.year * 12 + (d.month - 1) + months
year, month_index = divmod(total_months, 12)
month = month_index + 1
last_day = calendar.monthrange(year, month)[1]
day = min(d.day, last_day)
return datetime.date(year, month, day)
Turn #20Raw score: 1
147 in518 out419 reasoning64 cached665 total1.3s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write add_months(d: datetime.date, months: int) -> datetime.date to add months calendar months to d. months may be negative. If the day does not exist in the target month, use the last day of that month.
assistant
import calendar
import datetime
def add_months(d: datetime.date, months: int) -> datetime.date:
total_months = d.year * 12 + d.month - 1 + months
year, month_index = divmod(total_months, 12)
month = month_index + 1
day = min(d.day, calendar.monthrange(year, month)[1])
return datetime.date(year, month, day)
Turn #21Raw score: 1
147 in952 out844 reasoning128 cached1,099 total3.0s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write add_months(d: datetime.date, months: int) -> datetime.date to add months calendar months to d. months may be negative. If the day does not exist in the target month, use the last day of that month.
assistant
import calendar
import datetime
def add_months(d: datetime.date, months: int) -> datetime.date:
total_months = d.year * 12 + (d.month - 1) + months
year, month_index = divmod(total_months, 12)
month = month_index + 1
last_day = calendar.monthrange(year, month)[1]
day = min(d.day, last_day)
return datetime.date(year, month, day)
Turn #22Raw score: 1
171 in629 out539 reasoning64 cached800 total2.3s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write top_k_words(text: str, k: int) -> list[tuple[str, int]]. A word is a maximal sequence of ASCII letters and apostrophes. Words compare case-insensitively and are returned in lowercase. Return the k most frequent words as (word, count) tuples, sorted by highest count first, breaking ties alphabetically by word.
assistant
import re
from collections import Counter
def top_k_words(text: str, k: int) -> list[tuple[str, int]]:
if k <= 0:
return []
counts = Counter(match.lower() for match in re.findall(r"[A-Za-z']+", text))
return sorted(counts.items(), key=lambda item: (-item[1], item[0]))[:k]
Turn #23Raw score: 1
171 in873 out774 reasoning64 cached1,044 total2.5s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write top_k_words(text: str, k: int) -> list[tuple[str, int]]. A word is a maximal sequence of ASCII letters and apostrophes. Words compare case-insensitively and are returned in lowercase. Return the k most frequent words as (word, count) tuples, sorted by highest count first, breaking ties alphabetically by word.
assistant
import re
from collections import Counter
def top_k_words(text: str, k: int) -> list[tuple[str, int]]:
words = re.findall(r"[A-Za-z']+", text)
counts = Counter(word.lower() for word in words)
ordered = sorted(counts.items(), key=lambda item: (-item[1], item[0]))
if k <= 0:
return []
return ordered[:k]
Turn #24Raw score: 1
171 in927 out828 reasoning128 cached1,098 total2.6s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write top_k_words(text: str, k: int) -> list[tuple[str, int]]. A word is a maximal sequence of ASCII letters and apostrophes. Words compare case-insensitively and are returned in lowercase. Return the k most frequent words as (word, count) tuples, sorted by highest count first, breaking ties alphabetically by word.
assistant
import re
from collections import Counter
def top_k_words(text: str, k: int) -> list[tuple[str, int]]:
if k <= 0:
return []
words = re.findall(r"[A-Za-z']+", text)
counts = Counter(word.lower() for word in words)
ranked = sorted(counts.items(), key=lambda item: (-item[1], item[0]))
return ranked[:k]
Turn #25Raw score: 1
207 in2,450 out2,249 reasoning64 cached2,657 total7.2s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write flatten(obj: dict, sep: str = ".") -> dict. Flatten nested dicts and lists into a single level, joining keys with sep and using list indices as keys. Keep an empty dict or empty list as the value at its key. flatten({"a": {"b": 1}, "c": [{"d": 2}, 3], "e": {}}) returns {"a.b": 1, "c.0.d": 2, "c.1": 3, "e": {}}.
assistant
def flatten(obj: dict, sep: str = ".") -> dict:
result: dict = {}
def visit(value, prefix):
if isinstance(value, dict):
if not value:
if prefix is not None:
result[prefix] = {}
return
for key, item in value.items():
key_str = str(key)
next_prefix = key_str if prefix is None else f"{prefix}{sep}{key_str}"
visit(item, next_prefix)
elif isinstance(value, list):
if not value:
if prefix is not None:
result[prefix] = []
return
for index, item in enumerate(value):
next_prefix = str(index) if prefix is None else f"{prefix}{sep}{index}"
visit(item, next_prefix)
else:
if prefix is not None:
result[prefix] = value
visit(obj, None)
return result
Turn #26Raw score: 1
207 in1,872 out1,680 reasoning64 cached2,079 total4.8s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write flatten(obj: dict, sep: str = ".") -> dict. Flatten nested dicts and lists into a single level, joining keys with sep and using list indices as keys. Keep an empty dict or empty list as the value at its key. flatten({"a": {"b": 1}, "c": [{"d": 2}, 3], "e": {}}) returns {"a.b": 1, "c.0.d": 2, "c.1": 3, "e": {}}.
assistant
def flatten(obj: dict, sep: str = ".") -> dict:
result = {}
def walk(value, prefix):
if isinstance(value, dict):
if not value:
if prefix is not None:
result[prefix] = {}
return
for key, item in value.items():
next_prefix = str(key) if prefix is None else f"{prefix}{sep}{key}"
walk(item, next_prefix)
elif isinstance(value, list):
if not value:
if prefix is not None:
result[prefix] = []
return
for index, item in enumerate(value):
next_prefix = str(index) if prefix is None else f"{prefix}{sep}{index}"
walk(item, next_prefix)
else:
if prefix is not None:
result[prefix] = value
walk(obj, None)
return result
Turn #27Raw score: 1
207 in1,704 out1,515 reasoning192 cached1,911 total5.4s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write flatten(obj: dict, sep: str = ".") -> dict. Flatten nested dicts and lists into a single level, joining keys with sep and using list indices as keys. Keep an empty dict or empty list as the value at its key. flatten({"a": {"b": 1}, "c": [{"d": 2}, 3], "e": {}}) returns {"a.b": 1, "c.0.d": 2, "c.1": 3, "e": {}}.
assistant
def flatten(obj: dict, sep: str = ".") -> dict:
result = {}
def visit(value, prefix: str) -> None:
if isinstance(value, dict):
if not value:
if prefix:
result[prefix] = {}
return
for key, item in value.items():
child_prefix = str(key) if not prefix else f"{prefix}{sep}{key}"
visit(item, child_prefix)
elif isinstance(value, list):
if not value:
if prefix:
result[prefix] = []
return
for index, item in enumerate(value):
child_prefix = (
str(index) if not prefix else f"{prefix}{sep}{index}"
)
visit(item, child_prefix)
else:
if prefix:
result[prefix] = value
visit(obj, "")
return result
Turn #28Raw score: 0
177 in4,148 out4,012 reasoning64 cached4,325 total14.0s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write lru(maxsize: int), a decorator factory that caches a function's results by its positional arguments without using functools.lru_cache or functools.cache. It stores at most maxsize results. When the cache is full, evict the least recently used result. A cache hit counts as a use. The decorated function must preserve its original __name__.
assistant
def lru(maxsize: int):
def decorator(func):
cache = {}
def wrapper(*args, **kwargs):
key = args
if kwargs:
key = args + (tuple(sorted(kwargs.items())),)
if key in cache:
cache.move_to_end(key)
return cache[key]
result = func(*args, **kwargs)
cache[key] = result
if maxsize is not None:
while len(cache) > maxsize:
cache.popitem(last=False)
return result
wrapper.__name__ = func.__name__
return wrapper
return decorator
Turn #29Raw score: 1
177 in4,880 out4,719 reasoning64 cached5,057 total16.8s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write lru(maxsize: int), a decorator factory that caches a function's results by its positional arguments without using functools.lru_cache or functools.cache. It stores at most maxsize results. When the cache is full, evict the least recently used result. A cache hit counts as a use. The decorated function must preserve its original __name__.
assistant
def lru(maxsize):
def decorator(func):
cache = {}
def wrapper(*args):
if args in cache:
value = cache.pop(args)
cache[args] = value
return value
result = func(*args)
if maxsize is None or maxsize > 0:
if maxsize is not None and len(cache) >= maxsize:
del cache[next(iter(cache))]
cache[args] = result
return result
wrapper.__name__ = getattr(func, "__name__", wrapper.__name__)
wrapper.__qualname__ = getattr(func, "__qualname__", wrapper.__qualname__)
wrapper.__doc__ = func.__doc__
return wrapper
return decorator
Turn #30Raw score: 1
177 in3,196 out3,054 reasoning128 cached3,373 total11.1s
user
Write Python 3.13 code. Answer with one ```python code block that defines the function described below. Use only the standard library. Do not include explanations, examples, tests, input() calls, or print() calls.
Write lru(maxsize: int), a decorator factory that caches a function's results by its positional arguments without using functools.lru_cache or functools.cache. It stores at most maxsize results. When the cache is full, evict the least recently used result. A cache hit counts as a use. The decorated function must preserve its original __name__.
assistant
from collections import OrderedDict
from functools import wraps
def lru(maxsize):
def decorator(func):
cache = OrderedDict()
@wraps(func)
def wrapper(*args, **kwargs):
key = (args, tuple(sorted(kwargs.items())))
if key in cache:
cache.move_to_end(key)
return cache[key]
result = func(*args, **kwargs)
if maxsize is None or maxsize > 0:
if maxsize is not None and len(cache) >= maxsize:
cache.popitem(last=False)
cache[key] = result
return result
return wrapper
return decorator