leetcode-476 Number complement

 in the problem there given a positive integer num you need to convert in decimal to binary(0,1) let take num= 5 the binay convertion is 5(101) according to 8421 code we need complement of num is 010 and the integer  2 return its complement 

Input:5

Output:2

Explanation: the binary representation of 5 is 101 (no leading zero bits), and its compliment is 010. So you need to output 2

Input : 18

Output: 13

Explanation: The binary representation of 18 is 10010 (no leading zero bits), and its complement is 01101. so you need to output 13

The logic is:

without using complement the 5 binary code is 101 so we need convert to binary to decimal how to get 5, so like this the complement of 5 is also understand here there is any zero in binary code the 2 pow of any value is zero.

with complement of 5 is 2(010) so that is only main point to solve when ever there is zero only we do operation when there is one we no need for operation of 1 

This is the code in C++:

class Solution {

public:

    int findComplement(int num) {

        int sum=0;

        int i=0;

        while(num>0){

            if(num%2==0){

                sum+=pow(2,i)*1;

                }

            i++;

            num=num/2;

            }

        return sum;

    }

};

Explanation of the code:

we need sum of all converting values so int sum=0; it is we declared in while loop we converting num to binary  and also do operation where num%2 get the last value of the binary reperation so we need to 101 and by num=num/2to remove last value 10 and again the same proceess until the while loop break/false 10 

as i said what we neccessary that only we do if(num%2==0) this condition only sum and i++ is not conatant it increases until the condition of while false there is more important thing is i is started with zero that why  we declered int i=0; and the pow(2,i) is where the i=0 the binary of 5(101) the one is not equal to 0 so the condtion false no operation done and it reduced  num=num/2; so 10 and i increases to i=1 here binary(10) so there is condtion was true operation done (2^1)*1=>2 that added to sum and the binary reduced (1) then again the condition was false so again it reduced to 0 and while condition false the sum was returned 

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