Dourbt n leetcode problem

#1

class Solution {
public:
vector countBits(int num) {
vector res(num + 1);

    for(int i = 1; i < num + 1; i++) {
        if(i % 2 == 0) {
            res[i] = res[i >> 1];                
        }else{
            res[i] = res[i >> 1] + 1;
        }
    }
    
    return res;
}

};

In this problem,https://leetcode.com/problems/counting-bits/
if we are taking the num =5,if the coounting is taken from 1 then its accepted its giving a wrong ans for the count from 0 to 4 ,but if we dont take the cout from 0 then how we will get the 0 in the output list.Please explain me this I am getting confused

#2

hello @Somasree

why u stopped at 4 only ? here to get the count of bits in 5, u need to iterate till 5 na?

#3

for 0 answer is already 0 , by default vector will have 0 value thats why no need to take care of 0 explicitly

#4

Why we need to take 2 seperate cases for i=odd an di=even,because in this approach we are printing the decimal value of the binary

#5

…

we checking where the last bit (rightmost)is set or not.
for odd number , right most bit will be set. thats why if num is odd we are adding additional 1

#6

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