Permutations
Problem
Solution
We can generate permutations inductively. Suppose we have all the permutations
of nums[1:]
. If we prepend nums[1]
to these permutations, we now have
all permutations of nums
that start with nums[1]
. Doing this for each
\(n\) in nums
, we can generate every permutation of nums
.
Code
https://github.com/GeorgeRPu/tech-interview-prep/blob/main/solutions/Permutations.py
from typing import List
def permutations(nums: List[int]) -> List[List[int]]:
"""Generate all permutations of ``nums``.
"""
if len(nums) > 0:
return [[n] + p
for i, n in enumerate(nums)
for p in permutations(nums[:i] + nums[i + 1:])]
else:
return [[]]
Test
>>> from Permutations import permutations
>>> permutations([1, 2])
[[1, 2], [2, 1]]
>>> permutations([1, 2, 3])
[[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]
>>> len(permutations([1, 2, 3, 4]))
24
Functions
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Generate all permutations of |