Count number of 1's in binary representation

Question

Efficient way to count number of 1s in the binary representation of a number in O(1) if you have enough memory to play with. This is an interview question I found on an onli

Answer 1

I saw the following solution from another website:

int count_one(int x){
    x = (x & (0x55555555)) + ((x >> 1) & (0x55555555));
    x = (x & (0x33333333)) + ((x >> 2) & (0x33333333));
    x = (x & (0x0f0f0f0f)) + ((x >> 4) & (0x0f0f0f0f));
    x = (x & (0x00ff00ff)) + ((x >> 8) & (0x00ff00ff));
    x = (x & (0x0000ffff)) + ((x >> 16) & (0x0000ffff));
    return x;
}

Answer 2

   countBits(x){
     y=0;
     while(x){   
       y += x &  1 ;
       x  = x >> 1 ;
     }
   }

thats it?

Answer 3

I have actually done this using a bit of sleight of hand: a single lookup table with 16 entries will suffice and all you have to do is break the binary rep into nibbles (4-bit tuples). The complexity is in fact O(1) and I wrote a C++ template which was specialized on the size of the integer you wanted (in # bits)… makes it a constant expression instead of indetermined.

fwiw you can use the fact that (i & -i) will return you the LS one-bit and simply loop, stripping off the lsbit each time, until the integer is zero — but that’s an old parity trick.

Answer 4

The following is a C solution using bit operators:

int numberOfOneBitsInInteger(int input) {
  int numOneBits = 0;

  int currNum = input;
  while (currNum != 0) {
    if ((currNum & 1) == 1) {
      numOneBits++;
    }
    currNum = currNum >> 1;
  }
  return numOneBits;
}

The following is a Java solution using powers of 2:

public static int numOnesInBinary(int n) {

  if (n < 0) return -1;

  int j = 0;
  while ( n > Math.pow(2, j)) j++;

  int result = 0;
  for (int i=j; i >=0; i--){
    if (n >= Math.pow(2, i)) {
        n = (int) (n - Math.pow(2,i));
        result++;    
    }
  }

  return result;
}

Answer 5

Ruby implementation

def find_consecutive_1(n)
  num = n.to_s(2)
  arr = num.split("")
  counter = 0
  max = 0
  arr.each do |x|
      if x.to_i==1
          counter +=1
      else
          max = counter if counter > max
          counter = 0 
      end
      max = counter if counter > max  
  end
  max
end

puts find_consecutive_1(439)

Answer 6

Below are two simple examples (in C++) among many by which you can do this.

1. We can simply count set bits (1's) using __builtin_popcount().

   int numOfOnes(int x) { return __builtin_popcount(x); }

2. Loop through all bits in an integer, check if a bit is set and if it is then increment the count variable.

   int hammingDistance(int x) { int count = 0 for(int i = 0; i < 32; i++) if(x & (1 << i)) count++; return count; }

Hope this helps!