Day28. CountingBits.cpp

/*

Counting Bits

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 1:

Input: 2
Output: [0,1,1]
Example 2:

Input: 5
Output: [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.
   
   Hide Hint #1  
You should make use of what you have produced already.
   
   Hide Hint #2  
Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. And try to generate new range from previous.
   
   Hide Hint #3  
Or does the odd/even status of the number help you in calculating the number of 1s?

*/

class Solution {
public:
    vector<int> countBits(int num) {
        vector<int> sol(num+1,0);
        int next =2;
        int prev=0; 
        for(int i=1;i<=num;i++)
        {
            if(i%2!=0)  sol[i]=sol[i-1] +1;
            else 
            {
                if(i==next)
                {
                    sol[i]=1;
                    prev=next;
                    next=next*2;
                }
                else sol[i]=sol[prev]+sol[i-prev];
            }
        }
        return sol;
    }
};