Number of 1s in the two's complement binary representations of integers in a range

Question

This problem is from the 2011 Codesprint (http://csfall11.interviewstreet.com/):

One of the basics of Computer Science is knowing how numbers are represented in 2\'s

Answer 1

when a is positive, the better explanation was already been posted.

If a is negative, then on a 32-bit system each negative number between a and zero will have 32 1's bits less the number of bits in the range from 0 to the binary representation of positive a.

So, in a better way,

long long solve(int a) {
    if (a >= 0){
        if (a == 0) return 0;
        else if ((a %2) == 0) return solve(a - 1) + noOfSetBits(a);
        else return (2 * solve( a / 2)) + ((long long)a + 1) / 2;
    }else {
        a++;
        return ((long long)(-a) + 1) * 32 - solve(-a);
    }
}

Answer 2

Cast the array into a series of integers. Then for each integer do:

int NumberOfSetBits(int i)
{
   i = i - ((i >> 1) & 0x55555555);
   i = (i & 0x33333333) + ((i >> 2) & 0x33333333);
   return (((i + (i >> 4)) & 0x0F0F0F0F) * 0x01010101) >> 24;
}

Also this is portable, unlike __builtin_popcount

See here: How to count the number of set bits in a 32-bit integer?

Answer 3

Well, it's not that complicated...

The single-argument solve(int a) function is the key. It is short, so I will cut&paste it here:

long long solve(int a)
{
 if(a == 0) return 0 ;
 if(a % 2 == 0) return solve(a - 1) + __builtin_popcount(a) ;
 return ((long long)a + 1) / 2 + 2 * solve(a / 2) ;
}

It only works for non-negative a, and it counts the number of 1 bits in all integers from 0 to a inclusive.

The function has three cases:

a == 0 -> returns 0. Obviously.

a even -> returns the number of 1 bits in a plus solve(a-1). Also pretty obvious.

The final case is the interesting one. So, how do we count the number of 1 bits from 0 to an odd number a?

Consider all of the integers between 0 and a, and split them into two groups: The evens, and the odds. For example, if a is 5, you have two groups (in binary):

000  (aka. 0)
010  (aka. 2)
100  (aka. 4)

and

001  (aka 1)
011  (aka 3)
101  (aka 5)

Observe that these two groups must have the same size (because a is odd and the range is inclusive). To count how many 1 bits there are in each group, first count all but the last bits, then count the last bits.

All but the last bits looks like this:

00
01
10

...and it looks like this for both groups. The number of 1 bits here is just solve(a/2). (In this example, it is the number of 1 bits from 0 to 2. Also, recall that integer division in C/C++ rounds down.)

The last bit is zero for every number in the first group and one for every number in the second group, so those last bits contribute (a+1)/2 one bits to the total.

So the third case of the recursion is (a+1)/2 + 2*solve(a/2), with appropriate casts to long long to handle the case where a is INT_MAX (and thus a+1 overflows).

This is an O(log N) solution. To generalize it to solve(a,b), you just compute solve(b) - solve(a), plus the appropriate logic for worrying about negative numbers. That is what the two-argument solve(int a, int b) is doing.

Answer 4

In the following code, the bitsum of x is defined as the count of 1 bits in the two's complement representation of the numbers between 0 and x (inclusive), where Integer.MIN_VALUE <= x <= Integer.MAX_VALUE.

For example:

bitsum(0) is 0   
bitsum(1) is 1   
bitsum(2) is 1   
bitsum(3) is 4

..etc

10987654321098765432109876543210 i % 10 for 0 <= i <= 31
00000000000000000000000000000000 0
00000000000000000000000000000001 1
00000000000000000000000000000010 2
00000000000000000000000000000011 3
00000000000000000000000000000100 4
00000000000000000000000000000101 ...
00000000000000000000000000000110
00000000000000000000000000000111 (2^i)-1
00000000000000000000000000001000  2^i
00000000000000000000000000001001 (2^i)+1 
00000000000000000000000000001010 ...
00000000000000000000000000001011 x, 011 = x & (2^i)-1 = 3
00000000000000000000000000001100
00000000000000000000000000001101
00000000000000000000000000001110
00000000000000000000000000001111
00000000000000000000000000010000
00000000000000000000000000010001
00000000000000000000000000010010 18
...
01111111111111111111111111111111 Integer.MAX_VALUE

The formula of the bitsum is:

bitsum(x) = bitsum((2^i)-1) + 1 + x - 2^i + bitsum(x & (2^i)-1 )

Note that x - 2^i = x & (2^i)-1

Negative numbers are handled slightly differently than positive numbers. In this case the number of zeros is subtracted from the total number of bits:

Integer.MIN_VALUE <= x < -1
Total number of bits: 32 * -x.

The number of zeros in a negative number x is equal to the number of ones in -x - 1.

public class TwosComplement {
    //t[i] is the bitsum of (2^i)-1 for i in 0 to 31.
    private static long[] t = new long[32];
    static {
        t[0] = 0;
        t[1] = 1;
        int p = 2;
        for (int i = 2; i < 32; i++) {
            t[i] = 2*t[i-1] + p;
            p = p << 1;
        }
    }

    //count the bits between x and y inclusive
    public static long bitsum(int x, int y) {
        if (y > x && x > 0) {
            return bitsum(y) - bitsum(x-1);
        }
        else if (y >= 0 && x == 0) {
            return bitsum(y);
        }
        else if (y == x) {
            return Integer.bitCount(y);
        }
        else if (x < 0 && y == 0) {
            return bitsum(x);
        } else if (x < 0 && x < y && y < 0 ) {
            return bitsum(x) - bitsum(y+1);
        } else if (x < 0 && x < y && 0 < y) {
            return bitsum(x) + bitsum(y);
        }
        throw new RuntimeException(x + " " + y);
    }

    //count the bits between 0 and x
    public static long bitsum(int x) {
        if (x == 0) return 0;
        if (x < 0) {
            if (x == -1) {
                return 32;
            } else {
                long y = -(long)x;
                return 32 * y - bitsum((int)(y - 1));
            }
        } else {
            int n = x;
            int sum = 0;     //x & (2^i)-1
            int j = 0;
            int i = 1;       //i = 2^j
            int lsb = n & 1; //least significant bit
            n = n >>> 1;
            while (n != 0) {
                sum += lsb * i;
                lsb = n & 1;
                n = n >>> 1;
                i = i << 1;
                j++;
            }
            long tot = t[j] + 1 + sum + bitsum(sum);
            return tot;
        }
    }
}