best series encountered so far without using unneccessary logics - beginner friendly !!
@abhinavkumar77303 жыл бұрын
I remember I was struggling with the concept, your videos aided me a lot. Cheers....!
@CodingWithPrakash_3 жыл бұрын
Wow
@sanjana82672 жыл бұрын
Sir I have seriously struggled a lot with this stuff, thank you so much!
@zeta_meow_meow Жыл бұрын
nice and simple v
@desibhojan82876 ай бұрын
Good 👍😊
@DsaTraveller2 жыл бұрын
great playlist
@niteshgupta11182 жыл бұрын
Finally. Understood this one.
@khushichouksey253 Жыл бұрын
agr if we will continue when 2 comes again then how [1,2,2] will be formed?
@Rajmayank242 жыл бұрын
There will be no base case ??
@adarshupadhyay82653 жыл бұрын
What will be the time and space complexity of this solution?
@CodingWithPrakash_3 жыл бұрын
Watch subset video in this playlist for time and space complexity
@KalpitSharma2 жыл бұрын
Permutation k videos nahi dikh rhe
@theuntoldtree3 жыл бұрын
isme set vala trika thoda sa bta dete to sahi rehta
@theuntoldtree3 жыл бұрын
sir missed one major point , saying sort the array
@mohitkumar5572 жыл бұрын
if you sort the given array before any operation then the power set will be in the lexicographical order. That's it.
@sudiptodas62722 жыл бұрын
class Solution: def subsetsWithDup(self, nums): index=0 current=[] residual=[] nums.sort() return (self.subset(index,nums,current,residual)) def subset(self,index,nums,current,residual): current.sort() if current not in residual: temp=copy.deepcopy(current) residual.append(temp) for i in range(index,len(nums)): current.append(nums[i]) self.subset(i+1,nums,current,residual) current.pop(len(current)-1) return residual ** the filter logic is little complex to understand simple : sort the array b4 passing : nums.sort() before pushing to residual sort and check check if it is already present : current.sort()