i think the backtracking solution with decision tree is more intuitive. For each recursion call, you can store the number that has already added to the Answer list into the HashSet, and for each recursion call you can skip the number that already in the HashSet function permute(nums: number[]): number[][] { const result = [] const dfs = (set: Set, pm: number[]): void => { if (pm.length === nums.length) { result.push([...pm]) return } for (let i = 0; i < nums.length; i++) { if (set.has(nums[i])) { continue } pm.push(nums[i]) set.add(nums[i]) dfs(set, pm) pm.pop() set.delete(nums[i]) } } dfs(new Set(), []) return result }
@ashkan.arabim
29 күн бұрын
yeah I thought the same. he's also slicing and inserting in the middle of arrays which are O(n) operations.
@bobert6259
14 күн бұрын
I don't think I could've thought of this solution, the one I came up with was the backtracking one, so it's nice to learn a new approach at least! List copy is O(n) and slicing+inserting is also O(n). The backtracking approach only has a list copy (assuming appending and popping are O(1)). def permute(self, nums: List[int]) -> List[List[int]]: if not nums: return [] permutations = [] # To store all our permutations curr_indices = set() # To keep track of the indices used so far curr_path = [] # What stores the current permutation (temporarily) size = len(nums) # Store it for readability def dfs(idx): if len(curr_indices) == size: # Base case when we find a permutation permutations.append(curr_path.copy()) # Add a copy of the permutation return for i in range(size): # Go over every remaining possible addition if i not in curr_indices: # Constraint - do not explore already explored index curr_indices.add(i) # Keep track of added indices curr_path.append(nums[i]) # Store the updated permutation dfs((i + 1) % size) # Explore this new branch curr_indices.remove(i) # Once explored, remove it so this set can be used again curr_path.pop() # Same reasoning as removing from the set dfs(0) return permutations
@spiceybyte
7 күн бұрын
I came up with this direction as well, neetcode is usually great but I like this alternative solution better.
@dannygarcia7116
10 күн бұрын
Easy to understand (imo) DFS solution: class Solution: def permute(self, nums: List[int]) -> List[List[int]]: res = [] def dfs(perm): if len(perm) == len(nums): res.append(perm[:]) return for i in range(len(nums)): if nums[i] in perm: continue perm.append(nums[i]) dfs(perm) perm.pop() dfs([]) return res
@riddle-master-ruben
Ай бұрын
This is a way better explanation than the old video. I went months with this problem unsolved because I just couldn't understand, but this explanation helped me tremendously. Thanks!
@Gomeroff
Ай бұрын
I live in Russia, a new day has not yet begun, and you are already solving a new problem)
@pastori2672
Ай бұрын
i feel like a time traveler
@NguyenLe-nw2uj
14 күн бұрын
me too, i didn't look at the published timestamp, everything is kinda strange tho.
@bandarupawankumar7549
Ай бұрын
Thank you neetcode for again working on this problem :)
@AVGVSTVSivlivs
Ай бұрын
Swapping indices method for this question is probably more clear and you should also review Longest Palindromic Substring question. Solutions in the other questions are quite optimal. Thanks a lot!
@MyPodie
Ай бұрын
Morning Neetcode!
@johnveracruz2102
Ай бұрын
@NeetcodeIO PLS add a functionality on your site to not show the topics for the questions. want to do them in order without knowing the topic. thanks
@jitpatel7692
19 күн бұрын
You Make it easy, Thank you
@user-vu4ng4rb8k
Ай бұрын
hey also do video for cherry pickup 1 and bst to sorted DLL
@zereftribbiani8130
16 күн бұрын
I wouldn't recommend this way of doing it since slicing and inserting are O(n) operations
@kanjurer
Ай бұрын
I actually just solved the problem yesterday lmao
@ijavd
Ай бұрын
Morning
@pushkarsaini2
Ай бұрын
Now Solve 31 Next Permutation
@logn-x5e
27 күн бұрын
Ok, this one's really the first problem in neetcode150 that is actually VERY easy to understand the goal but I personally find it extremely hard to implement
@mukeshrawat1304
Ай бұрын
Is it possible for you to have some sort of poll or something where we could ask for a video solution of a leetcode problem which you haven't already solved. Because there are quite a few problems which don't have a proper solution on KZitem and God knows when they will appear as POTD.
@DeathSugar
Ай бұрын
I guess thats what his site is for.
@m_jdm357
Ай бұрын
The worst solution I've seen.
@NeetCodeIO
Ай бұрын
anything specific about it that you did not like?
@m_jdm357
Ай бұрын
@@NeetCodeIO I tried to understand it but it's really bad when debugging or writing down on paper. Really hard to understand. I found this solution: def permute(self, nums: List[int]) -> List[List[int]]: res = [] def backtrack(arr): if len(arr) == len(nums): res.append(arr.copy()) return for i in range(len(nums)): if nums[i] in arr: continue arr.append(nums[i]) backtrack(arr) arr.pop() backtrack([]) return res This I got. If no one supports me and everyone thinks your solution is good. Then I'm wrong.
@yunusemreozvarlik2906
19 күн бұрын
@@m_jdm357 Actually, I understood the concept better through this video and the solution that is being implemented is similar to yours. kzitem.info/news/bejne/yHyjl6B7n6CriIY
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