338.1 Bits 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:
For you num = 5 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 I do it in linear time O (n) /possibly in a single pass?
• Space complexity should be O (n).
• Can do it like a boss? Do it without the using any builtin function like __builtin_popcount in C + + or any other language.

Hint:

1. You should make the use of the produced already.
2. Divide the numbers in ranges like [2-3], [4-7], [8-15] and so on. and try to generate the new range from previous.
   
3. 
   

Law:

When I equals two of the power of N, the Hamming weight is 1.

Hammingweight (2) = 1

Hammingweight (3) = Hammingweight (2) + hammingweight (1) = 2

......

Hammingweight (8) = 1

Hammingweight (9) = Hammingweight (8) + hammingweight (1) = 2

Hammingweight (Ten) = Hammingweight (8) + hammingweight (2) = 2

Hammingweight (one) = Hammingweight (8) + hammingweight (2) = 3

  
 

   
  

  
static public int[] CountBits(int n)
   
  

  
 {
   
  

  
 int[] arr = new int[n + 1];
   
  

  
 arr[0] = 0;
   
  

  

   
  

  
 int pow2 = 2;
   
  

  
 int before = 1;
   
  

  
 for (int i = 1; i < n + 1; i++)
   
  

  
 {
   
  

  
 if (i == pow2)
   
  

  
 {
   
  

  
 arr[i] = 1;
   
  

  
 before = 1;
   
  

  
 pow2 <<= 1;
   
  

  
 }
   
  

  
 else
   
  

  
 {
   
  

  
 arr[i] = arr[before] + 1;
   
  

  
 before++;
   
  

  
 }
   
  

  
 }
   
  

  

   
  

  
 return arr;
   
  

  
 }
  
 

From for notes (Wiz)

338.1 Bits counting bits