@GopalGangoriya
00:04 Printing all subsets using power set technique 01:37 Power set is the number of subsets 03:16 Using simple Boolean logic to generate power set 04:54 Creating subsets using bit manipulation 06:27 Iterate and generate subsets using bit manipulation. 07:59 Generating power set using bit manipulation 09:34 Generating power set using bit manipulation. 11:14 Discussing space complexity and subset sizes in power set generation.
@arnabnath4782
No one can explain this with that ease..wonderful 👏🏼👏🏼👏🏼👏🏼
@santoshkumak
This problem looks very easy from explanation but after he writes code and mapping back to explanation you will know how tricky it is.... Kudos to you for putting this explanation and solution....
@rajdeep3493
watched it 3-4 times then i get the solution thanks man for explaining so nicely
@vishnupandey3390
nums=[1,2,3]
length=len(nums)
upper=(2**length)-1
sup_set=list()
for i in range(upper,-1,-1):
ans=list()
for j in range(length):
mask=1<<j
if i&mask : ans.append(nums[j])
sup_set.append(ans)
print(sup_set)
@AbhishekSharma-26
c++ code
class Solution {
public:
vector<vector<int>> subsets(vector<int>& nums) {
int n = nums.size();
int subsett = 1 << n;
vector<vector<int>> ans;
for(int num = 0; num < subsett; num++) {
vector<int> list;
for(int i = 0; i < n; i++) {
if(num & (1 << i)) {
list.push_back(nums[i]);
}
}
ans.push_back(list);
}
return ans;
}
};
@nishchaypandit973
Brother i have my placement in 3 months plz continue this playlist
@ssh3v
the observation that take set bit of 2^n index in input array as member of a subset is very theoretical & observation intensive and hard to come up with , if we have seen never before, how to deal with it @take U forward
@sarder681
striver its really deep knowledge .its hard to understand but code is simple and couple of lines.However understood boss.
@Nikhil-xx5bh
Take a bow 🏹 what an explanation 6:16
@dharanyuvi6951
Love this problem and the way of solving it
@unoshihaam
public static void main(String[] args) {
int[] nums = {1,2,3};
int subsets = 1<<nums.length;
ArrayList<ArrayList<Integer>> al = new ArrayList<>();
for(int i=0;i<subsets;i++){
ArrayList<Integer> subal = new ArrayList<>();
for(int j=0;j<nums.length;j++){
if((i&(1<<j))!=0){
subal.add(nums[j]);
}
}
al.add(subal);
}
System.out.println(al);
@jagadeesh6199
broo, i came after that recursion playlist only, there you are telling to watch bit manipulation playlist and here you are telling to watch recursion playlist🤭🤭😂🤣, btw you are the best thooo
@raghavmanish24
mjhe to khud pe pura beleive hai ki kv na soch pata mai ye .....mind blowing solution......thanku striver
@Tejs_45
I got the solution that how ur getting the subsequence but how is that final list ur storing in ans ? What data struc is that ans
@Cric_trick572
If(nums & (1<<i)) kounsa logic lga hai?
@lshuarckyma
the space complexity will be O(2^n) as 2^n is the sum of all the lengths of all the temp lists that will be created
@stith_pragya
Understood.....Thank You So Much for this wonderful video............🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻
@ritwikdutta6914
int n = nums.length;
List<List<Integer>> ans = new ArrayList<>();
for(int i = 0; i < (1 << n); i++){
int num = i;
ArrayList<Integer> ls = new ArrayList<>();
int j = 0;
while(num > 0){
if((num & 1) == 1)
ls.add(nums[j]);
j++;
num = num >> 1;
}
ans.add(ls);
}
return ans;
i did like this. In Java
@sachinvarma9949