L7. Single Number III | Bit Manipulation 

There is a small mistake @19:04 rightmost = (xor & (xor - 1)) ^ xor is the correct formula. Very great explanation! 🔥🙌

The previous way when you code at the end of the video was great. Please continue like that .

1:20: using hash map (brute force) 6:05: bitwise approach (optimised) 18:05: code for optimised approach

Great work again Striver! I wouldnt have ever tried learning DSA if i had never ran onto your channel. It is a request that you please make the string playlist next whenever you have free time(ik its hard to find time alongside doing a fulltime job,but you still do so much for us so thanks a lot).

The clearest explanation I've ever seen for 260 in bit manipulation. Thank you!

So nice of you Striver... THANK you so much for this type of content... Even words wouldn't be enough for that. Thanks a million.

at 13:00 by first bit he meant bit at index 1 from back (If any one was confused here)

C++ code with comments for better understanding class Solution { public: vector singleNumber(vector& nums) { // brute force using map // optimised using bit manipulation bucket // approach : we know that all duos xor will be 0 and the unique 2 elements xor will contain // nothing but a number which contain bits which are not same in both long long numXor=0; for(auto it:nums)numXor^=it; // now we need to distinguish both the numbers // and xor contains all the bits which are not same in both , we just need one different bit // taking the rightmost one is fine long long rightBit= numXor ^ (numXor &(numXor-1)); // now take 2 bucket integer which store numbers based on this rightBit int a=0,b=0; for(auto it:nums){ // if right bit is set if(it&rightBit)a^=it; else b^=it; } return {a,b}; } };

class Solution { public: vector singleNumber(vector& nums) { long num =0; for(int i=0;i

The best explanation ever. The previous one's last solution blew me away... Amazing Striver, done with Bit manipulation today. Will start greedy from Monday. Hope you sgtart the string and stacks queues playlist soon.

finally felt like old videos of striver are back

vector singleNumber(vector& nums) { long xorr = 0; //Taking xorr as long for the case of INT_MIN in nums[i] for(auto &it : nums) xorr ^= it; int rightMostBitTurnedOne = xorr & (-xorr); int b1 = 0, b2 = 0; //b1 --> 1st bit is set || b2 --> 1st bit is 0 for(auto &it : nums) { if(it & rightMostBitTurnedOne) b1 ^= it; else b2 ^= it; } return {b1, b2}; }

There is an mistake @19:04 it's not rightMost = (xor & (xor - 1)) & (xor); => rightMost = (xor & (xor - 1)) ^ (xor); or rightMost = xor & -xor is efficent way

here is my python3 code for the given problem 👇 nums=list(map(int,input("Enter the list : ").split(" "))) xor=0 for ele in nums: xor=xor^ele bitmask=xor&(xor-1) bitmask=bitmask^xor num1=0 num2=0 for ele in nums: if bitmask&ele: num1=num1^ele else: num2=num1^ele print(num1,"and",num2,"are the unique numbers out there in the list.")

(xor & (xor - 1)) ^ xor correction in formula

Understood........Thank You So Much for this wonderful video.............🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻🙏🏻

what if we get 4,8 both as unique no. appearing ones only , than they will go to the same bucket ...4^8 in bucket 2

Extremely helpful video. 5:54 just a small correction, there are two repeating numbers, hence m = n/2+2

Keep code at last as in your old videos ❤🎉

Legendary Explanation!!

How can we think off such solutions . There is only one method we have to learn this solution for this particular question ig.

U R the king striver !!

for rightmost xor&(-xor) will do I think

UNDERSTOOD;

19:06 there is a writing mistake the formula is rightmost= ((original-numb & (original-numb-1)^original-numb)

UNDERSTOOD Thank you so much sir

why are people asking for code?? It is literally the same as the pseudocode he has written.

understood

Thank you 😊

mind blowing logic

Can someone explain me how 14 and 4 got seperated and how we are checking rightmost bit pls do explain

Anyone is here that, it happens to me or others that the optimised approach is too tough using hashmap is simple, down and dusted.

XOR them as you get (naa hindi word ), ❤❤

thank you Bhaiya

Awesome!

Understood!

Thank u sir

thanks !

Understood

understood 🤩

public int singleNumber (int[]nums){ int xor=Arrays.stream(nums).reduce((a,b)->a^b).getAsInt(); int mask=xor& ~(xor-1); int[]ans=new int[2]; for(int num:nums){ if((num&mask)>0) ans[0]^=num; else ans[1]^=num; } return ans; } 🎉❤

9:40

This is goldddd

Once Again Make playlist or videos for Competitive programming because I have completed your dsa series 😊

13:28 "FUCK"

1st

understood

understood.