WEEK#6Dynamic Programming_Counting Bits

Description of the Problem

Given a non-negative integer number num. For every number 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 runtime 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.

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Solution1 (Mathematical pattern)

May R[i] = number of bits in i.
i.e. R[5] = 2 because binary(5) = (101)
We initial R[0] = 0, R[1] = 1, noticing such pattern :
R[2] = R[0] + 1;
R[3] = R[1] + 1;
R[4] = R[0] + 1;
R[5] = R[1] + 1;
R[6] = R[2] + 1;
R[7] = R[3] + 1;
R[8] = R[0] + 1;

image.png

From which we can code like this :

class Solution {
public:
    vector countBits(int num) {
        vector results;
        int size = num + 1;
        results.resize(size);
        results[0] = 0;
        results[1] = 1;
        for (int i = 2 ; i < size; i++)
            results[i] = results[i - exp2(floor(log2(i)))] + 1;
        return results;
    }
};

Efficiency.png

However, as can be seen, it's not the ideal solution

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Solution2

using bit operation, efficiency improved a great deal.

class Solution {
public:
    vector countBits(int num) {
        vector ret(num+1, 0);
        for (int i = 1; i <= num; ++i)
            ret[i] = ret[i&(i-1)] + 1;
        return ret;
    }
};

image.png