Count set bits in an integer using Lookup Table

Write an efficient program to count number of 1s in binary representation of an integer.
Examples: 

Input : n = 6
Output : 2
Binary representation of 6 is 110 and has 2 set bits

Input : n = 13
Output : 3
Binary representation of 13 is 1101 and has 3 set bits

In the previous post we had seen different method that solved this problem in O(log n) time. In this post we solve in O(1) using lookup table. Here we assume that the size of INT is 32-bits. It’s hard to count all 32 bits in one go using lookup table (” because it’s infeasible to create lookup table of size 2³²-1 “). So we break 32 bits into 8 bits of chunks( How lookup table of size (2⁸-1 ) index : 0-255 ).
LookUp Table 
In lookup tale, we store count of set_bit of every 
number that are in a range (0-255) 
LookupTable[0] = 0 | binary 00000000 CountSetBits 0 
LookupTable[1] = 1 | binary 00000001 CountSetBits 1 
LookupTable[2] = 1 | binary 00000010 CountSetBits 1 
LookupTable[3] = 2 | binary 00000011 CountSetBits 2 
LookupTable[4] = 1 | binary 00000100 CountSetBits 1 
and so...on upto LookupTable[255].
Let’s take an Example How lookup table work.

Let's number be : 354 
in Binary : 0000000000000000000000101100010

Split it into 8 bits chunks :
In Binary : 00000000 | 00000000 | 00000001 | 01100010
In decimal : 0 0 1 98

Now Count Set_bits using LookupTable
LookupTable[0] = 0
LookupTable[1] = 1
LookupTable[98] = 3

so Total bits count : 4 

 Below is the code implementation of the above approach:

Output:  

4 

Time Complexity: O(1)
Auxiliary Space: O(1)