Permutation Leetcode

Question

I was solving this leetcode permutation problem and came across an error that am getting n empty lists inside my returned list which suppose to print different permutations of the given list

getting output => [[], [], [], [], [], []]

Expected output=> [[1, 2, 3], [1, 3, 2], [2, 1, 3], [2, 3, 1], [3, 1, 2], [3, 2, 1]]

def permute(nums):
    l=[]
    s=list()
    ans=[]

        return helper(nums,s,l)
def helper(nums,s,l):
    if not nums:
        print(l)
        s.append(l)
    else:
        for i in range(len(nums)):
            c=nums[i]
            l.append(c)
            nums.pop(i)
            helper(nums,s,l)
            nums.insert(i,c)
            l.pop()
    return s
print(permute([1,2,3]))

sanyassh · Accepted Answer · 2019-09-15 21:05:37Z

7

You should do s.append(l.copy()) because otherwise you pop all values from the same list l, that's why the result consists of empty lists.

answered Sep 15, 2019 at 21:05

sanyassh

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4 Comments

yolon bsn Over a year ago

can you edit the code so that i could get the desired answer as am getting this [<function list.copy()>, <function list.copy()>, <function list.copy()>, <function list.copy()>, <function list.copy()>, <function list.copy()>]

sanyassh Over a year ago

@yolonbsn please check if you write exactly s.append(l.copy()), not s.append(l.copy) - note the parentheses.

yolon bsn Over a year ago

can you please tell me how the l is getting empty why did we used copy()

sanyassh Over a year ago

@yolonbsn it is getting empty because of two factors: 1) the list l is always the same in every call to helper function - it is passed by reference. 2) there is a line l.pop() to every line l.append(...) so in the end the number of calls to pop and append equals. To avoid this problem we use copy at the moment we need the value of l to be unchanged.

y051 · Accepted Answer · 2022-07-09 12:10:30Z

0

You can also use s.append(l[:]) this slicing also does the same

answered Jul 9, 2022 at 12:10

y051

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1 Comment

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