Find position of set bit gfg | GeeksforGeeks problem of the day | Bit Manipulation Basic

[@probabilitycodingisfunis1]

Code C++: if(N==0)return -1; if((N&N-1)==0)return log2(N)+1; return -1;

[@24_sanketgolar96]

if its having more than one ones in binary log2(N) will not be integer,it will be fraction,is it correct?

[@computerandenglish4896]

The algo or formula that you used is basically to find the power of 2. It is not to find the set bit. What if the given no is 7 i.e 111, then your algo would return 2 (which is not the answer). I would like to suggest a answer :- //if number is 0, then there is no set bit if (n == 0) { return -1; // -1 will represent that there is no set bit } //if number is other than 0, then int position = 1; while ((n & 1) != 1 ) { n = n >> 1; position++; //this variable stores the position of the set bit } return position; //here answer will be returned i.e. the position of the set bit

[@shivanshsri4072]

Very good explanation. Thank you soo much😊😊

[@jayeshchaurasiya262]

Love ur explanation ❣️

[@indiancseresearch6109]

can u plz make video on what is backtracking exactly ? how subset sum is backtracking problem ? when to apply it ? with examples . . . .

[@dhananjoydey1337]

Thanks

[@dhananjoydey1337]

Please upload today's problem of the day solution

[@rahuldwivedi9236]

BUT WHY?