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subset_generation.py 2.19 KB
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Maxim Smolskiy 提交于 2024-04-02 22:18 . Enable ruff RUF005 rule (#11344)
def subset_combinations(elements: list[int], n: int) -> list:
"""
Compute n-element combinations from a given list using dynamic programming.
Args:
elements: The list of elements from which combinations will be generated.
n: The number of elements in each combination.
Returns:
A list of tuples, each representing a combination of n elements.
>>> subset_combinations(elements=[10, 20, 30, 40], n=2)
[(10, 20), (10, 30), (10, 40), (20, 30), (20, 40), (30, 40)]
>>> subset_combinations(elements=[1, 2, 3], n=1)
[(1,), (2,), (3,)]
>>> subset_combinations(elements=[1, 2, 3], n=3)
[(1, 2, 3)]
>>> subset_combinations(elements=[42], n=1)
[(42,)]
>>> subset_combinations(elements=[6, 7, 8, 9], n=4)
[(6, 7, 8, 9)]
>>> subset_combinations(elements=[10, 20, 30, 40, 50], n=0)
[()]
>>> subset_combinations(elements=[1, 2, 3, 4], n=2)
[(1, 2), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)]
>>> subset_combinations(elements=[1, 'apple', 3.14], n=2)
[(1, 'apple'), (1, 3.14), ('apple', 3.14)]
>>> subset_combinations(elements=['single'], n=0)
[()]
>>> subset_combinations(elements=[], n=9)
[]
>>> from itertools import combinations
>>> all(subset_combinations(items, n) == list(combinations(items, n))
... for items, n in (
... ([10, 20, 30, 40], 2), ([1, 2, 3], 1), ([1, 2, 3], 3), ([42], 1),
... ([6, 7, 8, 9], 4), ([10, 20, 30, 40, 50], 1), ([1, 2, 3, 4], 2),
... ([1, 'apple', 3.14], 2), (['single'], 0), ([], 9)))
True
"""
r = len(elements)
if n > r:
return []
dp: list[list[tuple]] = [[] for _ in range(r + 1)]
dp[0].append(())
for i in range(1, r + 1):
for j in range(i, 0, -1):
for prev_combination in dp[j - 1]:
dp[j].append((*prev_combination, elements[i - 1]))
try:
return sorted(dp[n])
except TypeError:
return dp[n]
if __name__ == "__main__":
from doctest import testmod
testmod()
print(f"{subset_combinations(elements=[10, 20, 30, 40], n=2) = }")
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