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69 lines
1.8 KiB
Python
69 lines
1.8 KiB
Python
def kth_permutation(k: int, n: int) -> list[int]:
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"""
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Finds k'th lexicographic permutation (in increasing order) of
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0,1,2,...n-1 in O(n^2) time.
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:param k: The index of the permutation (0-based)
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:param n: The number of elements in the permutation
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:return: The k-th lexicographic permutation of size n
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Examples:
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First permutation is always 0,1,2,...n-1
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>>> kth_permutation(0, 5)
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[0, 1, 2, 3, 4]
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The order of permutation of 0,1,2,3 is [0,1,2,3], [0,1,3,2], [0,2,1,3],
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[0,2,3,1], [0,3,1,2], [0,3,2,1], [1,0,2,3], [1,0,3,2], [1,2,0,3],
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[1,2,3,0], [1,3,0,2]
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>>> kth_permutation(10, 4)
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[1, 3, 0, 2]
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>>> kth_permutation(10, 0)
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Traceback (most recent call last):
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...
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ValueError: n must be positive
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>>> kth_permutation(-1, 5)
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Traceback (most recent call last):
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...
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IndexError: k must be non-negative
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>>> kth_permutation(120, 5)
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Traceback (most recent call last):
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...
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IndexError: k out of bounds
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"""
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if n <= 0:
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raise ValueError("n must be positive")
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if k < 0:
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raise IndexError("k must be non-negative")
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# Factorials from 1! to (n-1)!
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factorials = [1]
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for i in range(2, n):
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factorials.append(factorials[-1] * i)
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if k >= factorials[-1] * n:
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raise IndexError("k out of bounds")
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permutation = []
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elements = list(range(n))
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# Find permutation
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while factorials:
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factorial = factorials.pop()
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number, k = divmod(k, factorial)
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permutation.append(elements[number])
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elements.pop(
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number
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) # elements.remove(elements[number]) is slower and redundant
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permutation.append(elements[0])
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return permutation
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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