Counting Bits

Description

Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1's in their binary representation and return them as an array.

Example: For num = 5 you should return [0,1,1,2,1,2].

Follow up:

• It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?

• Space complexity should be O(n).

• Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

Solutions

My solution

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> ans(num + 1);
        for(int i = 1; i <= num; ++i){
            if(i % 2)
                ans[i] = ans[i - 1] + 1;
            else{
                int count = ans[((i - 1) ^ i) >> 1];
                ans[i] = ans[i - 1] - count + 1;
            }
        }
        return ans;
    }
};

Optimal solution

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> ans(num + 1);
        for(int i = 1; i <= num; ++i){
            if(i % 2)
                ans[i] = ans[i >> 1] + 1; // or ans[i - 1] + 1;
            else
                ans[i] = ans[i >> 1];
        }
        return ans;
    }
};